Examkin

Differentiation of Trigonometric, Exponential, and Logarithmic Functions

24 soru

Tüm türler24Çoktan seçmeli13Sayısal11
Soru 1Soru →

If y=ln⁡(cos⁡2x)y = \ln(\cos 2x), what is dydx\frac{dy}{dx}?

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Cevap: −2tan⁡2x-2\tan 2x

Cevap

−2tan⁡2x-2\tan 2x
Applying the chain rule to y=ln⁡(cos⁡2x)y = \ln(\cos 2x) yields dydx=1cos⁡2x⋅(−sin⁡2x)⋅2=−2sin⁡2xcos⁡2x=−2tan⁡2x\frac{dy}{dx} = \frac{1}{\cos 2x} \cdot (-\sin 2x) \cdot 2 = -2\frac{\sin 2x}{\cos 2x} = -2\tan 2x.

Adım Adım Çözüm

1
Identify the inner and outer functions for the chain rule.
Let u=cos⁡2xu = \cos 2x, so y=ln⁡uy = \ln u.
The given function y=ln⁡(cos⁡2x)y = \ln(\cos 2x) is a composite transcendental function.
2
Differentiate yy with respect to uu, and uu with respect to xx.
dydu=1u=1cos⁡2x\frac{dy}{du} = \frac{1}{u} = \frac{1}{\cos 2x}, and dudx=−2sin⁡2x\frac{du}{dx} = -2\sin 2x.
The derivative of ln⁡u\ln u is 1u\frac{1}{u} and the derivative of cos⁡2x\cos 2x is −2sin⁡2x-2\sin 2x using the chain rule.
3
Apply the chain rule dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} and simplify.
dydx=1cos⁡2x⋅(−2sin⁡2x)=−2(sin⁡2xcos⁡2x)=−2tan⁡2x\frac{dy}{dx} = \frac{1}{\cos 2x} \cdot (-2\sin 2x) = -2\left(\frac{\sin 2x}{\cos 2x}\right) = -2\tan 2x.
Using the trigonometric identity sin⁡2xcos⁡2x=tan⁡2x\frac{\sin 2x}{\cos 2x} = \tan 2x simplifies the expression into standard form.

Anahtar Kavram

Differentiation of composite logarithmic and trigonometric functions using the Chain Rule
Tahmini Süre:1m 0s
Soru 2Soru →

If y=2e4x+ln⁡(2x+1)y = 2e^{4x} + \ln(2x + 1), what is the value of dydx\frac{dy}{dx} at x=0x = 0?

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Cevap: 10

Cevap

10
Differentiating y=2e4x+ln⁡(2x+1)y = 2e^{4x} + \ln(2x + 1) with respect to xx yields dydx=8e4x+22x+1\frac{dy}{dx} = 8e^{4x} + \frac{2}{2x + 1}. Evaluating this derivative at x=0x = 0 gives 8(e0)+22(0)+1=8(1)+2=108(e^0) + \frac{2}{2(0) + 1} = 8(1) + 2 = 10.

Adım Adım Çözüm

1
Differentiate each term of y=2e4x+ln⁡(2x+1)y = 2e^{4x} + \ln(2x + 1) with respect to xx
\frac{dy}{dx} = 8e^{4x} + \frac{2}{2x + 1}
By the chain rule, ddx(aekx)=akekx\frac{d}{dx}(ae^{kx}) = ak e^{kx} and ddx(ln⁡(u(x)))=u′(x)u(x)\frac{d}{dx}(\ln(u(x))) = \frac{u'(x)}{u(x)}.
2
Evaluate the derivative at x=0x = 0
\left.\frac{dy}{dx}\right|_{x=0} = 8e^{0} + \frac{2}{2(0) + 1} = 8(1) + \frac{2}{1} = 10
Substitute x=0x = 0 into the derived expression and simplify using e0=1e^0 = 1.

Anahtar Kavram

Differentiation of exponential and logarithmic functions using the chain rule
Soru 3Soru →

If y=e2xtan⁡xy = e^{2x} \tan x, what is dydx\frac{dy}{dx}?

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Cevap: e2x(2tan⁡x+sec⁡2x)e^{2x}(2\tan x + \sec^2 x)

Cevap

e2x(2tan⁡x+sec⁡2x)e^{2x}(2\tan x + \sec^2 x)
Applying the product rule ddx[uv]=u′v+uv′\frac{d}{dx}[uv] = u'v + uv' with u=e2xu = e^{2x} and v=tan⁡xv = \tan x gives u′=2e2xu' = 2e^{2x} and v′=sec⁡2xv' = \sec^2 x. Substituting these into the formula yields 2e2xtan⁡x+e2xsec⁡2x=e2x(2tan⁡x+sec⁡2x)2e^{2x}\tan x + e^{2x}\sec^2 x = e^{2x}(2\tan x + \sec^2 x).

Adım Adım Çözüm

1
Identify the component functions for the product rule
Let u(x)=e2xu(x) = e^{2x} and v(x)=tan⁡xv(x) = \tan x.
The given function y=e2xtan⁡xy = e^{2x} \tan x is a product of two functions.
2
Differentiate each component function separately
dudx=2e2x\frac{du}{dx} = 2e^{2x} by the chain rule, and dvdx=sec⁡2x\frac{dv}{dx} = \sec^2 x.
The derivative of ekxe^{kx} is kekxk e^{kx} and the derivative of tan⁡x\tan x is sec⁡2x\sec^2 x.
3
Apply the product rule formula dydx=udvdx+vdudx\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}
dydx=e2x⋅sec⁡2x+tan⁡x⋅2e2x\frac{dy}{dx} = e^{2x} \cdot \sec^2 x + \tan x \cdot 2e^{2x}.
Combining the terms according to the standard product rule.
4
Factor out the common term e2xe^{2x}
dydx=e2x(2tan⁡x+sec⁡2x)\frac{dy}{dx} = e^{2x}(2\tan x + \sec^2 x).
Simplifying to match standard exam response format.

Anahtar Kavram

Differentiation of Exponential and Trigonometric Functions using the Product Rule
Tahmini Süre:1m 30s
Soru 4Soru →

If y=ln⁡(1+sin⁡x1−sin⁡x)y = \ln\left(\frac{1 + \sin x}{1 - \sin x}\right), what is dydx\frac{dy}{dx}?

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Cevap: 2sec⁡x2\sec x

Cevap

The derivative dydx\frac{dy}{dx} is equal to 2sec⁡x2\sec x.
Rewriting the function as y=ln⁡(1+sin⁡x)−ln⁡(1−sin⁡x)y = \ln(1 + \sin x) - \ln(1 - \sin x) and differentiating both terms yields cos⁡x1+sin⁡x+cos⁡x1−sin⁡x\frac{\cos x}{1 + \sin x} + \frac{\cos x}{1 - \sin x}. Combining these over the common denominator (1−sin⁡2x)=cos⁡2x(1 - \sin^2 x) = \cos^2 x simplifies to 2cos⁡xcos⁡2x=2sec⁡x\frac{2\cos x}{\cos^2 x} = 2\sec x.

Adım Adım Çözüm

1
Apply the logarithmic law ln⁡(uv)=ln⁡u−ln⁡v\ln\left(\frac{u}{v}\right) = \ln u - \ln v
y=ln⁡(1+sin⁡x)−ln⁡(1−sin⁡x)y = \ln(1 + \sin x) - \ln(1 - \sin x)
Simplifies the quotient into separate terms prior to differentiation.
2
Differentiate each logarithmic term with respect to xx using the chain rule
dydx=11+sin⁡x⋅cos⁡x−11−sin⁡x⋅(−cos⁡x)=cos⁡x1+sin⁡x+cos⁡x1−sin⁡x\frac{dy}{dx} = \frac{1}{1 + \sin x} \cdot \cos x - \frac{1}{1 - \sin x} \cdot (-\cos x) = \frac{\cos x}{1 + \sin x} + \frac{\cos x}{1 - \sin x}
The derivative of ln⁡(f(x))\ln(f(x)) is f′(x)f(x)\frac{f'(x)}{f(x)}.
3
Combine the fractions over a common denominator
dydx=cos⁡x(1−sin⁡x)+cos⁡x(1+sin⁡x)(1+sin⁡x)(1−sin⁡x)=2cos⁡x1−sin⁡2x\frac{dy}{dx} = \frac{\cos x (1 - \sin x) + \cos x (1 + \sin x)}{(1 + \sin x)(1 - \sin x)} = \frac{2\cos x}{1 - \sin^2 x}
Adding two fractions with denominators (1+sin⁡x)(1 + \sin x) and (1−sin⁡x)(1 - \sin x).
4
Use the Pythagorean trigonometric identity 1−sin⁡2x=cos⁡2x1 - \sin^2 x = \cos^2 x and simplify
dydx=2cos⁡xcos⁡2x=2cos⁡x=2sec⁡x\frac{dy}{dx} = \frac{2\cos x}{\cos^2 x} = \frac{2}{\cos x} = 2\sec x
Simplifying 2cos⁡xcos⁡2x\frac{2\cos x}{\cos^2 x} yields 2sec⁡x2\sec x.

Anahtar Kavram

Differentiation of Logarithmic and Trigonometric Functions via Chain Rule and Log Properties

Alternatif Yöntem

Alternatively, express y=ln⁡(sec⁡x+tan⁡x)2=2ln⁡(sec⁡x+tan⁡x)y = \ln(\sec x + \tan x)^2 = 2 \ln(\sec x + \tan x). The derivative of ln⁡(sec⁡x+tan⁡x)\ln(\sec x + \tan x) is sec⁡xtan⁡x+sec⁡2xsec⁡x+tan⁡x=sec⁡x\frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} = \sec x. Multiplying by 2 gives 2sec⁡x2\sec x.
Tahmini Süre:2m 0s
Soru 5Soru →

If y=e4x−sin⁡xy = e^{4x} - \sin x, what is dydx\frac{dy}{dx}?

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Cevap: 4e4x−cos⁡x4e^{4x} - \cos x

Cevap

4e4x−cos⁡x4e^{4x} - \cos x
Differentiating e4xe^{4x} gives 4e4x4e^{4x} by applying the chain rule, and differentiating −sin⁡x-\sin x yields −cos⁡x-\cos x. Combining these terms gives the correct derivative 4e4x−cos⁡x4e^{4x} - \cos x.

Adım Adım Çözüm

1
Apply the sum/difference rule of differentiation.
\frac{dy}{dx} = \frac{d}{dx}(e^{4x}) - \frac{d}{dx}(\sin x)
The derivative of a difference of two terms is the difference of their individual derivatives.
2
Differentiate the exponential term e4xe^{4x} using the chain rule.
ddx(e4x)=4e4x\frac{d}{dx}(e^{4x}) = 4e^{4x}
By the chain rule, \frac{d}{dx}(e^{k x}) = k e^{k x}.
3
Differentiate the trigonometric term sin⁡x\sin x.
ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x
The standard derivative of sin⁡x\sin x with respect to xx is cos⁡x\cos x.
4
Combine the results.
\frac{dy}{dx} = 4e^{4x} - \cos x
Subtracting the derivative of sin⁡x\sin x from the derivative of e4xe^{4x} gives the final answer.

Anahtar Kavram

Differentiation of Exponential and Trigonometric Functions
Soru 6Soru →

If y=5sin⁡x+e3xy = 5\sin x + e^{3x}, what is the value of dydx\frac{dy}{dx} at x=0x = 0?

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Cevap: 8

Cevap

The value of the derivative at x=0x = 0 is 8.
Differentiating y=5sin⁡x+e3xy = 5\sin x + e^{3x} yields dydx=5cos⁡x+3e3x\frac{dy}{dx} = 5\cos x + 3e^{3x}. Substituting x=0x = 0 gives 5cos⁡(0)+3e0=5(1)+3(1)=85\cos(0) + 3e^{0} = 5(1) + 3(1) = 8.

Adım Adım Çözüm

1
Differentiate each term of the function y=5sin⁡x+e3xy = 5\sin x + e^{3x} with respect to xx.
\frac{dy}{dx} = 5\cos x + 3e^{3x}
The derivative of sin⁡x\sin x is cos⁡x\cos x, and applying the chain rule to e3xe^{3x} gives 3e3x3e^{3x}.
2
Evaluate the derivative dydx\frac{dy}{dx} at x=0x = 0.
5\cos(0) + 3e^0 = 5(1) + 3(1) = 8
Evaluating trigonometric and exponential functions at zero gives cos⁡(0)=1\cos(0) = 1 and e0=1e^0 = 1.

Anahtar Kavram

Differentiation of trigonometric and exponential functions and evaluation of derivatives at specific points.
Soru 7Soru →

If y=e2xcos⁡3xy = e^{2x} \cos 3x, what is dydx\frac{dy}{dx}?

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Cevap: e2x(2cos⁡3x−3sin⁡3x)e^{2x}(2\cos 3x - 3\sin 3x)

Cevap

dydx=e2x(2cos⁡3x−3sin⁡3x)\frac{dy}{dx} = e^{2x}(2\cos 3x - 3\sin 3x)
Applying the product rule to u=e2xu = e^{2x} and v=cos⁡3xv = \cos 3x yields dudx=2e2x\frac{du}{dx} = 2e^{2x} and dvdx=−3sin⁡3x\frac{dv}{dx} = -3\sin 3x. Substituting these into udvdx+vdudxu\frac{dv}{dx} + v\frac{du}{dx} gives e2x(−3sin⁡3x)+cos⁡3x(2e2x)=e2x(2cos⁡3x−3sin⁡3x)e^{2x}(-3\sin 3x) + \cos 3x(2e^{2x}) = e^{2x}(2\cos 3x - 3\sin 3x).

Adım Adım Çözüm

1
Identify the components for the product rule
Let u=e2xu = e^{2x} and v=cos⁡3xv = \cos 3x.
The function y=e2xcos⁡3xy = e^{2x} \cos 3x is a product of two differentiable functions.
2
Differentiate u=e2xu = e^{2x} with respect to xx
dudx=2e2x\frac{du}{dx} = 2e^{2x}
By the chain rule, ddx(ekx)=kekx\frac{d}{dx}(e^{kx}) = k e^{kx}.
3
Differentiate v=cos⁡3xv = \cos 3x with respect to xx
dvdx=−3sin⁡3x\frac{dv}{dx} = -3\sin 3x
By the chain rule, ddx(cos⁡kx)=−ksin⁡kx\frac{d}{dx}(\cos kx) = -k \sin kx.
4
Apply the product rule formula dydx=udvdx+vdudx\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} and factor out e2xe^{2x}
\frac{dy}{dx} = e^{2x}(-3\sin 3x) + (\cos 3x)(2e^{2x}) = e^{2x}(2\cos 3x - 3\sin 3x)
Combining the products and factoring out the common exponential factor simplifies the expression.

Anahtar Kavram

Product Rule and Chain Rule for Exponential and Trigonometric Functions
Soru 8Soru →

If y=ln⁡(e3xcos⁡x)y = \ln(e^{3x} \cos x), what is dydx\frac{dy}{dx}?

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Cevap: 3−tan⁡x3 - \tan x

Cevap

3−tan⁡x3 - \tan x
Applying the product rule for logarithms converts y=ln⁡(e3xcos⁡x)y = \ln(e^{3x} \cos x) into y=3x+ln⁡(cos⁡x)y = 3x + \ln(\cos x). Differentiating 3x3x yields 33, and differentiating ln⁡(cos⁡x)\ln(\cos x) by chain rule yields −sin⁡xcos⁡x=−tan⁡x\frac{-\sin x}{\cos x} = -\tan x. Combining these gives 3−tan⁡x3 - \tan x.

Adım Adım Çözüm

1
Simplify the logarithmic function using log laws
y=ln⁡(e3x)+ln⁡(cos⁡x)=3x+ln⁡(cos⁡x)y = \ln(e^{3x}) + \ln(\cos x) = 3x + \ln(\cos x)
Splitting the logarithm of a product simplifies the differentiation process.
2
Differentiate each term with respect to xx
\frac{dy}{dx} = \frac{d}{dx}(3x) + \frac{d}{dx}(\ln(\cos x)) = 3 + \frac{1}{\cos x} \cdot (-\sin x)
The derivative of 3x3x is 33, and applying the chain rule to ln⁡(u)\ln(u) gives 1u⋅dudx\frac{1}{u} \cdot \frac{du}{dx} where u=cos⁡xu = \cos x.
3
Simplify the trigonometric ratio
\frac{dy}{dx} = 3 - \frac{\sin x}{\cos x} = 3 - \tan x
Replacing sin⁡xcos⁡x\frac{\sin x}{\cos x} with tan⁡x\tan x yields the simplified final derivative.

Anahtar Kavram

Differentiation of Logarithmic and Trigonometric Functions via Chain Rule and Log Laws
Soru 9Soru →

If y=cos⁡(3x)+exy = \cos(3x) + e^x, what is dydx\frac{dy}{dx}?

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Cevap: −3sin⁡(3x)+ex-3\sin(3x) + e^x

Cevap

−3sin⁡(3x)+ex-3\sin(3x) + e^x
Differentiating y=cos⁡(3x)+exy = \cos(3x) + e^x term-by-term yields dydx=−3sin⁡(3x)+ex\frac{dy}{dx} = -3\sin(3x) + e^x. By the chain rule, ddxcos⁡(3x)=−3sin⁡(3x)\frac{d}{dx}\cos(3x) = -3\sin(3x), while the derivative of exe^x is simply exe^x.

Adım Adım Çözüm

1
Differentiate the trigonometric term cos⁡(3x)\cos(3x) using the chain rule
ddx(cos⁡(3x))=−sin⁡(3x)⋅ddx(3x)=−3sin⁡(3x)\frac{d}{dx}(\cos(3x)) = -\sin(3x) \cdot \frac{d}{dx}(3x) = -3\sin(3x)
The derivative of cos⁡(u)\cos(u) is −sin⁡(u)⋅u′-\sin(u) \cdot u'
2
Differentiate the exponential term exe^x
ddx(ex)=ex\frac{d}{dx}(e^x) = e^x
The exponential function exe^x is its own derivative
3
Combine the results using the sum rule
dydx=−3sin⁡(3x)+ex\frac{dy}{dx} = -3\sin(3x) + e^x
The derivative of a sum of functions is the sum of their individual derivatives

Anahtar Kavram

Differentiation of trigonometric and exponential functions using the chain rule
Soru 10Soru →

If y=ln⁡(sec⁡3x+tan⁡3x)y = \ln(\sec 3x + \tan 3x), what is dydx\frac{dy}{dx}?

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Cevap: 3sec⁡3x3\sec 3x

Cevap

3sec⁡3x3\sec 3x
Using the chain rule for y=ln⁡uy = \ln u where u=sec⁡3x+tan⁡3xu = \sec 3x + \tan 3x, we find u′=3sec⁡3xtan⁡3x+3sec⁡23x=3sec⁡3x(tan⁡3x+sec⁡3x)u' = 3\sec 3x\tan 3x + 3\sec^2 3x = 3\sec 3x(\tan 3x + \sec 3x). Dividing u′u' by uu cancels out (sec⁡3x+tan⁡3x)(\sec 3x + \tan 3x), leaving 3sec⁡3x3\sec 3x.

Adım Adım Çözüm

1
Apply the chain rule for logarithmic functions ddx[ln⁡(u)]=1ududx\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx}.
Set u=sec⁡3x+tan⁡3xu = \sec 3x + \tan 3x, so dydx=1sec⁡3x+tan⁡3x⋅ddx(sec⁡3x+tan⁡3x)\frac{dy}{dx} = \frac{1}{\sec 3x + \tan 3x} \cdot \frac{d}{dx}(\sec 3x + \tan 3x).
The function is a composite function of the form y=ln⁡(u(x))y = \ln(u(x)).
2
Differentiate the inner function u=sec⁡3x+tan⁡3xu = \sec 3x + \tan 3x using the chain rule.
\frac{du}{dx} = 3\sec 3x \tan 3x + 3\sec^2 3x.
The derivative of sec⁡(ax)\sec(ax) is asec⁡(ax)tan⁡(ax)a\sec(ax)\tan(ax) and the derivative of tan⁡(ax)\tan(ax) is asec⁡2(ax)a\sec^2(ax).
3
Factor out common terms in the numerator and simplify the expression.
\frac{dy}{dx} = \frac{3\sec 3x(\tan 3x + \sec 3x)}{\sec 3x + \tan 3x} = 3\sec 3x.
The term (tan⁡3x+sec⁡3x)(\tan 3x + \sec 3x) in the numerator cancels with the denominator (sec⁡3x+tan⁡3x)(\sec 3x + \tan 3x).

Anahtar Kavram

Differentiation of Logarithmic and Trigonometric Functions via the Chain Rule
Tahmini Süre:1m 30s
Soru 11Soru →

If y=e3xsin⁡(2x)y = e^{3x} \sin(2x), what is dydx\frac{dy}{dx}?

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Cevap: e3x(3sin⁡2x+2cos⁡2x)e^{3x}(3\sin 2x + 2\cos 2x)

Cevap

e3x(3sin⁡2x+2cos⁡2x)e^{3x}(3\sin 2x + 2\cos 2x)
Using the product rule dydx=udvdx+vdudx\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} on y=e3xsin⁡(2x)y = e^{3x}\sin(2x) gives e3x⋅2cos⁡(2x)+sin⁡(2x)⋅3e3xe^{3x} \cdot 2\cos(2x) + \sin(2x) \cdot 3e^{3x}. Factoring out e3xe^{3x} results in e3x(3sin⁡2x+2cos⁡2x)e^{3x}(3\sin 2x + 2\cos 2x).

Adım Adım Çözüm

1
Identify the function components for the product rule
Let u=e3xu = e^{3x} and v=sin⁡(2x)v = \sin(2x).
The function yy is a product of two differentiable functions u(x)u(x) and v(x)v(x).
2
Differentiate each component using the chain rule
dudx=3e3x\frac{du}{dx} = 3e^{3x} and dvdx=2cos⁡(2x)\frac{dv}{dx} = 2\cos(2x).
Applying the chain rule gives ddx(e3x)=3e3x\frac{d}{dx}(e^{3x}) = 3e^{3x} and ddx(sin⁡2x)=2cos⁡2x\frac{d}{dx}(\sin 2x) = 2\cos 2x.
3
Apply the product rule formula dydx=udvdx+vdudx\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}
dydx=e3x(2cos⁡2x)+sin⁡(2x)(3e3x)=e3x(3sin⁡2x+2cos⁡2x)\frac{dy}{dx} = e^{3x}(2\cos 2x) + \sin(2x)(3e^{3x}) = e^{3x}(3\sin 2x + 2\cos 2x).
Combining the products and factoring out the common exponential factor e3xe^{3x} yields the final derivative.

Anahtar Kavram

Differentiation of Transcendental Functions using Product Rule and Chain Rule
Tahmini Süre:1m 30s
Soru 12Soru →

If y=esin⁡2xln⁡(tan⁡x)y = e^{\sin 2x} \ln(\tan x), what is the value of dydx\frac{dy}{dx} at x=π4x = \frac{\pi}{4}?

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Cevap: 2e2e

Cevap

The derivative evaluated at x=π4x = \frac{\pi}{4} is 2e2e.
Applying the product rule dydx=u′v+uv′\frac{dy}{dx} = u'v + uv' to y=esin⁡2xln⁡(tan⁡x)y = e^{\sin 2x} \ln(\tan x) gives u′=2cos⁡(2x)esin⁡2xu' = 2\cos(2x)e^{\sin 2x} and v′=2csc⁡(2x)v' = 2\csc(2x). At x=π4x = \frac{\pi}{4}, cos⁡(2x)=cos⁡(π/2)=0\cos(2x) = \cos(\pi/2) = 0, which vanishes the first term. The second term evaluates to esin⁡(π/2)⋅2csc⁡(π/2)=e1⋅2(1)=2ee^{\sin(\pi/2)} \cdot 2\csc(\pi/2) = e^1 \cdot 2(1) = 2e.

Adım Adım Çözüm

1
Identify the product rule structure for y=u⋅vy = u \cdot v, where u=esin⁡2xu = e^{\sin 2x} and v=ln⁡(tan⁡x)v = \ln(\tan x).
The product rule states dydx=u′v+uv′\frac{dy}{dx} = u'v + uv'.
The given function is a product of an exponential function and a logarithmic function.
2
Differentiate u=esin⁡2xu = e^{\sin 2x} using the chain rule.
dudx=ddx(sin⁡2x)⋅esin⁡2x=2cos⁡(2x)esin⁡2x\frac{du}{dx} = \frac{d}{dx}(\sin 2x) \cdot e^{\sin 2x} = 2\cos(2x) e^{\sin 2x}.
The inner function is sin⁡2x\sin 2x, whose derivative is 2cos⁡2x2\cos 2x.
3
Differentiate v=ln⁡(tan⁡x)v = \ln(\tan x) using the chain rule.
dvdx=1tan⁡x⋅sec⁡2x=cos⁡xsin⁡x⋅1cos⁡2x=1sin⁡xcos⁡x=2csc⁡(2x)\frac{dv}{dx} = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = 2\csc(2x).
Simplifying sec⁡2xtan⁡x\frac{\sec^2 x}{\tan x} gives 2sin⁡2x=2csc⁡(2x)\frac{2}{\sin 2x} = 2\csc(2x).
4
Combine terms into the product rule expression.
dydx=[2cos⁡(2x)esin⁡2xln⁡(tan⁡x)]+[esin⁡2x⋅2csc⁡(2x)]\frac{dy}{dx} = [2\cos(2x) e^{\sin 2x} \ln(\tan x)] + [e^{\sin 2x} \cdot 2\csc(2x)].
Substitute u,u′,v,v′u, u', v, v' into dydx=u′v+uv′\frac{dy}{dx} = u'v + uv'.
5
Evaluate the derivative at x=π4x = \frac{\pi}{4}.
At x=π4x = \frac{\pi}{4}, 2x=π22x = \frac{\pi}{2}. Thus, cos⁡(π/2)=0\cos(\pi/2) = 0, tan⁡(π/4)=1  ⟹  ln⁡(1)=0\tan(\pi/4) = 1 \implies \ln(1) = 0, sin⁡(π/2)=1\sin(\pi/2) = 1, and csc⁡(π/2)=1\csc(\pi/2) = 1. The first term becomes 00, and the second term becomes e1⋅2(1)=2ee^{1} \cdot 2(1) = 2e. Thus, dydx=2e\frac{dy}{dx} = 2e.
Substitute special angle values into the expression to compute the numerical result.

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Product Rule and Chain Rule for Transcendental Functions
Soru 13Soru →

If y=3e2x+ln⁡(cos⁡x)y = 3e^{2x} + \ln(\cos x), calculate the value of dydx\frac{dy}{dx} at x=0x = 0.

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Cevap: 6

Cevap

The value of the derivative of y=3e2x+ln⁡(cos⁡x)y = 3e^{2x} + \ln(\cos x) at x=0x = 0 is 6.
Differentiating each transcendental term individually using the chain rule yields dydx=6e2x−tan⁡x\frac{dy}{dx} = 6e^{2x} - \tan x. Substituting x=0x = 0 gives 6e0−tan⁡0=6−0=66e^0 - \tan 0 = 6 - 0 = 6.

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1
Differentiate the exponential term 3e2x3e^{2x}
6e2x6e^{2x}
Applying the derivative rule for exponential functions ddx[aekx]=akekx\frac{d}{dx}[a e^{kx}] = a k e^{kx}.
2
Differentiate the logarithmic term ln⁡(cos⁡x)\ln(\cos x)
−tan⁡x-\tan x
Applying the chain rule ddx[ln⁡(u)]=1ududx\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx} where u=cos⁡xu = \cos x gives −sin⁡xcos⁡x=−tan⁡x\frac{-\sin x}{\cos x} = -\tan x.
3
Combine terms and evaluate at x=0x = 0
6
Substituting x=0x = 0 into dydx=6e2x−tan⁡x\frac{dy}{dx} = 6e^{2x} - \tan x yields 6e0−tan⁡0=6(1)−0=66e^0 - \tan 0 = 6(1) - 0 = 6.

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Differentiation of Trigonometric, Exponential, and Logarithmic Functions
Soru 14Soru →

If y=2sin⁡(3x)+e4xy = 2\sin(3x) + e^{4x}, what is the value of dydx\frac{dy}{dx} at x=0x = 0?

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Cevap: 10

Cevap

10
Differentiating y=2sin⁡(3x)+e4xy = 2\sin(3x) + e^{4x} yields dydx=6cos⁡(3x)+4e4x\frac{dy}{dx} = 6\cos(3x) + 4e^{4x}. Evaluating this expression at x=0x = 0 gives 6cos⁡(0)+4e0=6(1)+4(1)=106\cos(0) + 4e^{0} = 6(1) + 4(1) = 10.

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1
Differentiate each term of y=2sin⁡(3x)+e4xy = 2\sin(3x) + e^{4x} with respect to xx.
dydx=6cos⁡(3x)+4e4x\frac{dy}{dx} = 6\cos(3x) + 4e^{4x}
Applying the chain rule gives ddx[2sin⁡(3x)]=2×3cos⁡(3x)=6cos⁡(3x)\frac{d}{dx}[2\sin(3x)] = 2 \times 3\cos(3x) = 6\cos(3x) and ddx[e4x]=4e4x\frac{d}{dx}[e^{4x}] = 4e^{4x}.
2
Evaluate the derivative at x=0x = 0.
6\cos(0) + 4e^{0} = 6(1) + 4(1) = 10
Substituting x=0x = 0 gives cos⁡(0)=1\cos(0) = 1 and e0=1e^{0} = 1.

Anahtar Kavram

Differentiation of trigonometric and exponential functions
Soru 15Soru →

If y=e2xln⁡(ex+sin⁡x)y = e^{2x} \ln(e^x + \sin x), determine the value of d2ydx2\frac{d^2 y}{dx^2} at x=0x = 0.

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Cevap: 5

Cevap

5
Evaluating the second derivative of y=e2xln⁡(ex+sin⁡x)y = e^{2x} \ln(e^x + \sin x) at x=0x = 0 yields 55. This is calculated by applying the product, chain, and quotient rules to get d2ydx2\frac{d^2 y}{dx^2}, and substituting x=0x = 0, where e0=1e^0 = 1, sin⁡0=0\sin 0 = 0, cos⁡0=1\cos 0 = 1, and ln⁡1=0\ln 1 = 0.

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1
Differentiate y=e2xln⁡(ex+sin⁡x)y = e^{2x} \ln(e^x + \sin x) with respect to xx using the product rule.
dydx=2e2xln⁡(ex+sin⁡x)+e2x(ex+cos⁡xex+sin⁡x)\frac{dy}{dx} = 2e^{2x} \ln(e^x + \sin x) + e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right)
The function is composed of u(x)=e2xu(x) = e^{2x} and v(x)=ln⁡(ex+sin⁡x)v(x) = \ln(e^x + \sin x). By chain rule, v′(x)=ex+cos⁡xex+sin⁡xv'(x) = \frac{e^x + \cos x}{e^x + \sin x}.
2
Differentiate the first term T1(x)=2e2xln⁡(ex+sin⁡x)T_1(x) = 2e^{2x} \ln(e^x + \sin x) to get T1′(x)T_1'(x).
T1′(x)=4e2xln⁡(ex+sin⁡x)+2e2x(ex+cos⁡xex+sin⁡x)T_1'(x) = 4e^{2x} \ln(e^x + \sin x) + 2e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right)
Applying the product rule to 2e2x2e^{2x} and ln⁡(ex+sin⁡x)\ln(e^x + \sin x).
3
Differentiate the second term T2(x)=e2x(ex+cos⁡xex+sin⁡x)T_2(x) = e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right) using the product and quotient rules.
T2′(x)=2e2x(ex+cos⁡xex+sin⁡x)+e2x((ex−sin⁡x)(ex+sin⁡x)−(ex+cos⁡x)2(ex+sin⁡x)2)T_2'(x) = 2e^{2x} \left(\frac{e^x + \cos x}{e^x + \sin x}\right) + e^{2x} \left(\frac{(e^x - \sin x)(e^x + \sin x) - (e^x + \cos x)^2}{(e^x + \sin x)^2}\right)
The derivative of ex+cos⁡xex+sin⁡x\frac{e^x + \cos x}{e^x + \sin x} requires the quotient rule.
4
Evaluate T1′(0)T_1'(0) and T2′(0)T_2'(0) at x=0x = 0.
T1′(0)=4(1)(0)+2(1)(2)=4T_1'(0) = 4(1)(0) + 2(1)(2) = 4, and T2′(0)=2(1)(2)+1((1)(1)−2212)=4−3=1T_2'(0) = 2(1)(2) + 1 \left(\frac{(1)(1) - 2^2}{1^2}\right) = 4 - 3 = 1.
At x=0x = 0, e0=1e^0 = 1, sin⁡0=0\sin 0 = 0, cos⁡0=1\cos 0 = 1, and ln⁡(1)=0\ln(1) = 0.
5
Sum the evaluated derivative components to find d2ydx2∣x=0\frac{d^2 y}{dx^2}\Big|_{x=0}.
d2ydx2∣x=0=T1′(0)+T2′(0)=4+1=5\frac{d^2 y}{dx^2}\Big|_{x=0} = T_1'(0) + T_2'(0) = 4 + 1 = 5
Combining the evaluated terms gives the final numerical value.

Anahtar Kavram

Higher-Order Derivatives of Combined Transcendental Functions
Soru 16Soru →

If y=5e2x+cos⁡(4x)y = 5e^{2x} + \cos(4x), find the value of dydx\frac{dy}{dx} at x=0x = 0.

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Cevap: 10

Cevap

10
Differentiating y=5e2x+cos⁡(4x)y = 5e^{2x} + \cos(4x) with respect to xx yields dydx=10e2x−4sin⁡(4x)\frac{dy}{dx} = 10e^{2x} - 4\sin(4x). Substituting x=0x = 0 gives 10e0−4sin⁡(0)=10(1)−0=1010e^0 - 4\sin(0) = 10(1) - 0 = 10.

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1
Differentiate y=5e2x+cos⁡(4x)y = 5e^{2x} + \cos(4x) with respect to xx.
dydx=10e2x−4sin⁡(4x)\frac{dy}{dx} = 10e^{2x} - 4\sin(4x)
The derivative of eaxe^{ax} is aeaxa e^{ax} and the derivative of cos⁡(ax)\cos(ax) is −asin⁡(ax)-a \sin(ax).
2
Evaluate the derivative at x=0x = 0.
10
Substitute x=0x = 0 into 10e2x−4sin⁡(4x)10e^{2x} - 4\sin(4x) to obtain 10(1)−4(0)=1010(1) - 4(0) = 10.

Anahtar Kavram

Differentiation of exponential and trigonometric functions
Soru 17Soru →

If y=e2xcos⁡xy = \frac{e^{2x}}{\cos x}, calculate the value of the derivative dydx\frac{dy}{dx} at x=0x = 0.

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Cevap: 2

Cevap

The value of the derivative dydx\frac{dy}{dx} at x=0x = 0 is 2.
Applying the quotient rule to y=e2xcos⁡xy = \frac{e^{2x}}{\cos x} yields dydx=2e2xcos⁡x+e2xsin⁡xcos⁡2x\frac{dy}{dx} = \frac{2e^{2x}\cos x + e^{2x}\sin x}{\cos^2 x}. Substituting x=0x = 0 gives 2(1)(1)+(1)(0)12=2\frac{2(1)(1) + (1)(0)}{1^2} = 2.

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1
Set up the quotient rule components for y=u(x)v(x)y = \frac{u(x)}{v(x)}.
Let u(x)=e2xu(x) = e^{2x} and v(x)=cos⁡xv(x) = \cos x.
The given function is a ratio of exponential and trigonometric functions.
2
Compute individual derivatives dudx\frac{du}{dx} and dvdx\frac{dv}{dx}.
dudx=2e2x\frac{du}{dx} = 2e^{2x} and dvdx=−sin⁡x\frac{dv}{dx} = -\sin x.
Using the chain rule for exponential functions and standard derivative rules for trigonometric functions.
3
Apply the quotient rule formula dydx=vu′−uv′v2\frac{dy}{dx} = \frac{v u' - u v'}{v^2}.
dydx=cos⁡x(2e2x)−e2x(−sin⁡x)cos⁡2x=e2x(2cos⁡x+sin⁡x)cos⁡2x\frac{dy}{dx} = \frac{\cos x (2e^{2x}) - e^{2x}(-\sin x)}{\cos^2 x} = \frac{e^{2x}(2\cos x + \sin x)}{\cos^2 x}.
Combining terms gives the exact derivative function.
4
Evaluate the derivative at x=0x = 0.
dydx∣x=0=e0(2cos⁡0+sin⁡0)cos⁡20=1⋅(2+0)1=2\frac{dy}{dx}\Big|_{x=0} = \frac{e^{0}(2\cos 0 + \sin 0)}{\cos^2 0} = \frac{1 \cdot (2 + 0)}{1} = 2.
Using trigonometric and exponential values at zero: e0=1e^0 = 1, cos⁡0=1\cos 0 = 1, and sin⁡0=0\sin 0 = 0.

Anahtar Kavram

Differentiation of Exponential and Trigonometric Functions using Quotient Rule
Soru 18Soru →

If y=esin⁡2(3x)y = e^{\sin^2(3x)}, what is dydx\frac{dy}{dx}?

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Cevap: 3sin⁡(6x)esin⁡2(3x)3\sin(6x)e^{\sin^2(3x)}

Cevap

The derivative dydx\frac{dy}{dx} is 3sin⁡(6x)esin⁡2(3x)3\sin(6x)e^{\sin^2(3x)}.
Differentiating y=esin⁡2(3x)y = e^{\sin^2(3x)} requires applying the chain rule step-by-step: first differentiating the exponential function to get esin⁡2(3x)e^{\sin^2(3x)}, then differentiating sin⁡2(3x)\sin^2(3x) to obtain 2sin⁡(3x)⋅3cos⁡(3x)=6sin⁡(3x)cos⁡(3x)2\sin(3x) \cdot 3\cos(3x) = 6\sin(3x)\cos(3x). Multiplying these together and applying the double-angle identity 2sin⁡(3x)cos⁡(3x)=sin⁡(6x)2\sin(3x)\cos(3x) = \sin(6x) yields 3sin⁡(6x)esin⁡2(3x)3\sin(6x)e^{\sin^2(3x)}.

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1
Identify the inner function uu and outer function yy for applying the chain rule.
Let u=sin⁡2(3x)=(sin⁡(3x))2u = \sin^2(3x) = (\sin(3x))^2, so y=euy = e^u.
The function is an exponential function whose exponent is a composite trigonometric function.
2
Differentiate u=(sin⁡(3x))2u = (\sin(3x))^2 with respect to xx using the chain rule.
\frac{du}{dx} = 2\sin(3x) \cdot \frac{d}{dx}(\sin(3x)) = 2\sin(3x) \cdot 3\cos(3x) = 6\sin(3x)\cos(3x).
The derivative of [g(x)]2[g(x)]^2 is 2g(x)g′(x)2g(x)g'(x), and ddx(sin⁡(3x))=3cos⁡(3x)\frac{d}{dx}(\sin(3x)) = 3\cos(3x).
3
Differentiate y=euy = e^u with respect to xx using dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.
dydx=esin⁡2(3x)⋅6sin⁡(3x)cos⁡(3x).\frac{dy}{dx} = e^{\sin^2(3x)} \cdot 6\sin(3x)\cos(3x).
The derivative of eue^u with respect to uu is eue^u.
4
Simplify the expression using the trigonometric double-angle identity 2sin⁡θcos⁡θ=sin⁡(2θ)2\sin\theta\cos\theta = \sin(2\theta), where θ=3x\theta = 3x.
\frac{dy}{dx} = 3 \cdot (2\sin(3x)\cos(3x)) e^{\sin^2(3x)} = 3\sin(6x)e^{\sin^2(3x)}.
Rewriting 6sin⁡(3x)cos⁡(3x)6\sin(3x)\cos(3x) as 3sin⁡(6x)3\sin(6x) simplifies the expression to standard examination form.

Anahtar Kavram

Chain Rule for Composite Exponential and Trigonometric Functions
Tahmini Süre:2m 0s
Soru 19Soru →

If y=ln⁡(cos⁡(3x))y = \ln(\cos(3x)), what is dydx\frac{dy}{dx}?

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Cevap: −3tan⁡(3x)-3\tan(3x)

Cevap

−3tan⁡(3x)-3\tan(3x)
Differentiating y=ln⁡(cos⁡(3x))y = \ln(\cos(3x)) requires using the chain rule twice: first for ln⁡(u)\ln(u), giving 1cos⁡(3x)\frac{1}{\cos(3x)}, and then for cos⁡(3x)\cos(3x), giving −3sin⁡(3x)-3\sin(3x). Multiplying these results yields −3sin⁡(3x)cos⁡(3x)\frac{-3\sin(3x)}{\cos(3x)}, which simplifies to −3tan⁡(3x)-3\tan(3x).

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1
Apply the chain rule formula for differentiating logarithmic functions ddx[ln⁡(u)]=1ududx\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx}
dydx=1cos⁡(3x)⋅ddx(cos⁡(3x))\frac{dy}{dx} = \frac{1}{\cos(3x)} \cdot \frac{d}{dx}(\cos(3x))
The outer function is ln⁡(u)\ln(u) where u=cos⁡(3x)u = \cos(3x).
2
Differentiate the inner trigonometric function cos⁡(3x)\cos(3x)
ddx(cos⁡(3x))=−3sin⁡(3x)\frac{d}{dx}(\cos(3x)) = -3\sin(3x)
By the chain rule, ddx[cos⁡(kx)]=−ksin⁡(kx)\frac{d}{dx}[\cos(kx)] = -k\sin(kx).
3
Multiply the terms and simplify using trigonometric identities
dydx=−3sin⁡(3x)cos⁡(3x)=−3tan⁡(3x)\frac{dy}{dx} = \frac{-3\sin(3x)}{\cos(3x)} = -3\tan(3x)
Using the identity sin⁡(θ)cos⁡(θ)=tan⁡(θ)\frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta), the derivative simplifies to −3tan⁡(3x)-3\tan(3x).

Anahtar Kavram

Differentiation of Logarithmic and Trigonometric Functions via Chain Rule
Tahmini Süre:1m 30s
Soru 20Soru →

If y=e−xtan⁡x+ln⁡(2x+1)y = e^{-x} \tan x + \ln(2x + 1), find the value of dydx\frac{dy}{dx} at x=0x = 0.

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Cevap: 3

Cevap

The value of dydx\frac{dy}{dx} at x=0x = 0 is 3.
Applying the product rule to e−xtan⁡xe^{-x}\tan x gives −e−xtan⁡x+e−xsec⁡2x-e^{-x}\tan x + e^{-x}\sec^2 x, and applying the chain rule to ln⁡(2x+1)\ln(2x+1) gives 22x+1\frac{2}{2x+1}. Evaluating dydx=−e−xtan⁡x+e−xsec⁡2x+22x+1\frac{dy}{dx} = -e^{-x}\tan x + e^{-x}\sec^2 x + \frac{2}{2x+1} at x=0x = 0 yields −e0(0)+e0(1)+21=3-e^0(0) + e^0(1) + \frac{2}{1} = 3.

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1
Differentiate u(x)=e−xtan⁡xu(x) = e^{-x} \tan x using the product rule.
\frac{du}{dx} = -e^{-x} \tan x + e^{-x} \sec^2 x
The derivative of e−xe^{-x} is −e−x-e^{-x} and the derivative of tan⁡x\tan x is \sec^2 x.
2
Differentiate v(x)=ln⁡(2x+1)v(x) = \ln(2x + 1) using the chain rule.
dvdx=22x+1\frac{dv}{dx} = \frac{2}{2x + 1}
The derivative of ln⁡(g(x))\ln(g(x)) is g′(x)g(x)\frac{g'(x)}{g(x)}, where g(x)=2x+1g(x) = 2x + 1 and g′(x)=2g'(x) = 2.
3
Sum the derivatives to find the complete expression for dydx\frac{dy}{dx}.
\frac{dy}{dx} = -e^{-x} \tan x + e^{-x} \sec^2 x + \frac{2}{2x + 1}
The derivative of a sum is equal to the sum of the derivatives.
4
Evaluate the derivative at x=0x = 0.
\frac{dy}{dx}\Big|_{x=0} = -e^0(0) + e^0(1)^2 + \frac{2}{1} = 3
Since tan⁡(0)=0\tan(0) = 0, e0=1e^0 = 1, and sec⁡(0)=1\sec(0) = 1, substituting x=0x = 0 simplifies the derivative to 0+1+2=30 + 1 + 2 = 3.

Anahtar Kavram

Differentiation of Transcendental Functions (Product and Chain Rules)
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Differentiation of Trigonometric, Exponential, and Logarithmic Functions Alıştırma Soruları — JAMB UTME | Examkin