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Differentiation of Trigonometric, Exponential, and Logarithmic Functions

24 questions

All types24Multiple choice13Numeric11
Question 21Question →

If y=e3xcos⁡(2x)+ln⁡(x+1)y = e^{3x}\cos(2x) + \ln(x + 1), calculate the value of dydx\frac{dy}{dx} at x=0x = 0.

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Answer: 4

Answer

The derivative evaluated at x=0x = 0 is 4.
Differentiating y=e3xcos⁡(2x)+ln⁡(x+1)y = e^{3x}\cos(2x) + \ln(x + 1) with respect to xx yields dydx=3e3xcos⁡(2x)−2e3xsin⁡(2x)+1x+1\frac{dy}{dx} = 3e^{3x}\cos(2x) - 2e^{3x}\sin(2x) + \frac{1}{x+1}. Evaluating this derivative at x=0x = 0 gives 3(1)(1)−2(1)(0)+1=43(1)(1) - 2(1)(0) + 1 = 4.

Step-by-Step Solution

1
Differentiate the product u(x)=e3xcos⁡(2x)u(x) = e^{3x}\cos(2x) using the product rule and chain rule.
dudx=3e3xcos⁡(2x)−2e3xsin⁡(2x)\frac{du}{dx} = 3e^{3x}\cos(2x) - 2e^{3x}\sin(2x)
By the product rule ddx(uv)=u′v+uv′\frac{d}{dx}(uv) = u'v + uv', where ddx(e3x)=3e3x\frac{d}{dx}(e^{3x}) = 3e^{3x} and ddx(cos⁡(2x))=−2sin⁡(2x)\frac{d}{dx}(\cos(2x)) = -2\sin(2x).
2
Differentiate the logarithmic term v(x)=ln⁡(x+1)v(x) = \ln(x + 1).
dvdx=1x+1\frac{dv}{dx} = \frac{1}{x + 1}
The derivative of ln⁡(g(x))\ln(g(x)) is g′(x)g(x)\frac{g'(x)}{g(x)}.
3
Combine the terms to write the complete derivative dydx\frac{dy}{dx}.
\frac{dy}{dx} = 3e^{3x}\cos(2x) - 2e^{3x}\sin(2x) + \frac{1}{x + 1}
The derivative of a sum of functions is the sum of their individual derivatives.
4
Evaluate dydx\frac{dy}{dx} at x=0x = 0.
\frac{dy}{dx}\Big|_{x=0} = 3e^0\cos(0) - 2e^0\sin(0) + \frac{1}{0 + 1} = 3(1)(1) - 2(1)(0) + 1 = 4
Substitute x=0x = 0 using e0=1e^0 = 1, cos⁡(0)=1\cos(0) = 1, and sin⁡(0)=0\sin(0) = 0.

Key Concept

Differentiation of Trigonometric, Exponential, and Logarithmic Functions
Estimated Time:1m 30s
Question 22Question →

If y=e3xcos⁡x+sin⁡xy = \frac{e^{3x}}{\cos x + \sin x}, what is the value of dydx\frac{dy}{dx} at x=0x = 0?

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Answer: 2

Answer

The value of dydx\frac{dy}{dx} at x=0x = 0 is 22.
Applying the quotient rule dydx=vu′−uv′v2\frac{dy}{dx} = \frac{v u' - u v'}{v^2} to y=e3xcos⁡x+sin⁡xy = \frac{e^{3x}}{\cos x + \sin x} yields dydx=3e3x(cos⁡x+sin⁡x)−e3x(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)2\frac{dy}{dx} = \frac{3e^{3x}(\cos x + \sin x) - e^{3x}(\cos x - \sin x)}{(\cos x + \sin x)^2}. Evaluating this expression at x=0x = 0 gives 3(1)(1)−(1)(1)12=2\frac{3(1)(1) - (1)(1)}{1^2} = 2.

Step-by-Step Solution

1
Identify numerator u(x)u(x) and denominator v(x)v(x) for the quotient rule.
u(x)=e3xu(x) = e^{3x} and v(x)=cos⁡x+sin⁡xv(x) = \cos x + \sin x.
The function yy is expressed as a quotient of exponential and trigonometric functions.
2
Find the first derivatives of u(x)u(x) and v(x)v(x).
u′(x)=3e3xu'(x) = 3e^{3x} and v′(x)=−sin⁡x+cos⁡xv'(x) = -\sin x + \cos x.
Derivative of ekxe^{kx} is kekxk e^{kx}, derivative of cos⁡x\cos x is −sin⁡x-\sin x, and derivative of sin⁡x\sin x is cos⁡x\cos x.
3
Substitute u(x)u(x), v(x)v(x), u′(x)u'(x), and v′(x)v'(x) into the quotient rule formula dydx=u′v−uv′v2\frac{dy}{dx} = \frac{u'v - uv'}{v^2}.
dydx=3e3x(cos⁡x+sin⁡x)−e3x(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)2\frac{dy}{dx} = \frac{3e^{3x}(\cos x + \sin x) - e^{3x}(\cos x - \sin x)}{(\cos x + \sin x)^2}.
The quotient rule is required to differentiate u(x)v(x)\frac{u(x)}{v(x)}.
4
Evaluate the derivative expression at x=0x = 0.
dydx∣x=0=3(1)(1+0)−1(1−0)(1+0)2=3−11=2\frac{dy}{dx}\Big|_{x=0} = \frac{3(1)(1 + 0) - 1(1 - 0)}{(1 + 0)^2} = \frac{3 - 1}{1} = 2.
At x=0x = 0, e0=1e^0 = 1, cos⁡0=1\cos 0 = 1, and sin⁡0=0\sin 0 = 0.

Key Concept

Differentiation of exponential and trigonometric functions using the quotient rule
Question 23Question →

If y=e−2xsin⁡(3x)y = e^{-2x} \sin(3x), find dydx\frac{dy}{dx}.

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

Answer

dydx=e−2x(3cos⁡(3x)−2sin⁡(3x))\frac{dy}{dx} = e^{-2x} (3\cos(3x) - 2\sin(3x))
Applying the product rule ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} to u=e−2xu = e^{-2x} and v=sin⁡(3x)v = \sin(3x) yields dudx=−2e−2x\frac{du}{dx} = -2e^{-2x} and dvdx=3cos⁡(3x)\frac{dv}{dx} = 3\cos(3x). Substituting these terms gives e−2x(3cos⁡(3x))+sin⁡(3x)(−2e−2x)=e−2x(3cos⁡(3x)−2sin⁡(3x))e^{-2x}(3\cos(3x)) + \sin(3x)(-2e^{-2x}) = e^{-2x}(3\cos(3x) - 2\sin(3x)).

Step-by-Step Solution

1
Identify the component functions for the product rule
Let u=e−2xu = e^{-2x} and v=sin⁡(3x)v = \sin(3x).
The function yy is a product of an exponential function and a trigonometric function.
2
Differentiate each component using the chain rule
dudx=−2e−2x\frac{du}{dx} = -2e^{-2x} and dvdx=3cos⁡(3x)\frac{dv}{dx} = 3\cos(3x).
ddx(ekx)=kekx\frac{d}{dx}(e^{kx}) = k e^{kx} and ddx(sin⁡(kx))=kcos⁡(kx)\frac{d}{dx}(\sin(kx)) = k \cos(kx) where kk is a constant.
3
Apply the product rule formula dydx=udvdx+vdudx\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} and factor out e−2xe^{-2x}
dydx=e−2x(3cos⁡(3x))+sin⁡(3x)(−2e−2x)=e−2x(3cos⁡(3x)−2sin⁡(3x))\frac{dy}{dx} = e^{-2x}(3\cos(3x)) + \sin(3x)(-2e^{-2x}) = e^{-2x}(3\cos(3x) - 2\sin(3x)).
Combining terms correctly gives the exact derivative.

Key Concept

Differentiation of Product of Transcendental Functions (Exponential and Trigonometric)
Question 24Question →

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

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Answer: 5.5

Answer

The value of dydx\frac{dy}{dx} at x=0x = 0 is 5.55.5.
Differentiating ln⁡(2+sin⁡(3x))\ln(2 + \sin(3x)) by the chain rule gives 3cos⁡(3x)2+sin⁡(3x)\frac{3\cos(3x)}{2 + \sin(3x)}, and differentiating e4xe^{4x} gives 4e4x4e^{4x}. Evaluating 3cos⁡(3x)2+sin⁡(3x)+4e4x\frac{3\cos(3x)}{2 + \sin(3x)} + 4e^{4x} at x=0x = 0 yields 3(1)2+0+4(1)=1.5+4=5.5\frac{3(1)}{2 + 0} + 4(1) = 1.5 + 4 = 5.5.

Step-by-Step Solution

1
Differentiate the logarithmic component ln⁡(2+sin⁡(3x))\ln(2 + \sin(3x)) using the chain rule
ddx[ln⁡(2+sin⁡(3x))]=3cos⁡(3x)2+sin⁡(3x)\frac{d}{dx}[\ln(2 + \sin(3x))] = \frac{3\cos(3x)}{2 + \sin(3x)}
By the chain rule, ddx[ln⁡(u)]=1ududx\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx}, where u=2+sin⁡(3x)u = 2 + \sin(3x) and dudx=3cos⁡(3x)\frac{du}{dx} = 3\cos(3x).
2
Differentiate the exponential component e4xe^{4x}
ddx[e4x]=4e4x\frac{d}{dx}[e^{4x}] = 4e^{4x}
The standard rule for exponential differentiation states that ddx[ekx]=kekx\frac{d}{dx}[e^{kx}] = k e^{kx}.
3
Combine the results to state the derivative function dydx\frac{dy}{dx}
dydx=3cos⁡(3x)2+sin⁡(3x)+4e4x\frac{dy}{dx} = \frac{3\cos(3x)}{2 + \sin(3x)} + 4e^{4x}
The derivative of a sum of functions is the sum of their individual derivatives.
4
Evaluate dydx\frac{dy}{dx} at x=0x = 0
dydx∣x=0=3cos⁡(0)2+sin⁡(0)+4e0=3(1)2+0+4(1)=1.5+4=5.5\frac{dy}{dx}\Big|_{x=0} = \frac{3\cos(0)}{2 + \sin(0)} + 4e^0 = \frac{3(1)}{2 + 0} + 4(1) = 1.5 + 4 = 5.5
Substituting x=0x = 0 uses the values cos⁡(0)=1\cos(0) = 1, sin⁡(0)=0\sin(0) = 0, and e0=1e^0 = 1.

Key Concept

Differentiation of Logarithmic, Trigonometric, and Exponential Functions using the Chain Rule
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