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Zorluk: KolayDifferentiation of Trigonometric, Exponential, and Logarithmic Functions

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

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.

Adım Adım Çözüm

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
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