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Worked solution

Combine chain and product rules with elementary functions

Differentiate with respect to xx: (a) 4e2x+14e^{2x+1}; (b) x2sin⁡xx^2\sin x; (c) ln⁡(3x)\ln(3x).

CJC · O Level revision · Example 28

(a) Differentiate the exponential and its exponent

Answer

8e2x+18e^{2x+1}.

Multiply by the inner derivative two

ddx(4e2x+1)=4e2x+1ddx(2x+1)=4e2x+1⋅2=8e2x+1\begin{aligned}&\frac d{dx}(4e^{2x+1})\\ & =4e^{2x+1}\frac d{dx}(2x+1)\\ & =4e^{2x+1}\cdot2\\ & =8e^{2x+1}\end{aligned}

The constant 4 remains while the derivative of 2x+1 supplies 2.

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(b) Use the product rule with sine in radians

Answer

x(xcos⁡x+2sin⁡x)x(x\cos x+2\sin x).

Differentiate both factors and factor x

ddx(x2sin⁡x)=x2ddx(sin⁡x)+sin⁡xddx(x2)=x2cos⁡x+2xsin⁡x=x(xcos⁡x+2sin⁡x)\begin{aligned}&\frac d{dx}(x^2\sin x)\\ & =x^2\frac d{dx}(\sin x)+\sin x\frac d{dx}(x^2)\\ & =x^2\cos x+2x\sin x\\ & =x(x\cos x+2\sin x)\end{aligned}

The trigonometric calculus rule assumes radians. Retain the product-rule stage before simplifying.

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(c) Differentiate the logarithm on its domain

Answer

1/x1/x, for x>0x>0.

Divide by the argument and multiply by its derivative

ddxln⁡(3x)=13xddx(3x)=13x⋅3=1x\begin{aligned}&\frac d{dx}\ln(3x)\\ & =\frac1{3x}\frac d{dx}(3x)\\ & =\frac1{3x}\cdot3\\ & =\frac1x\end{aligned}

The original ln⁡(3x)\ln(3x) needs 3x>03x>0, hence x>0. Simplifying the derivative does not extend the original logarithm to negative inputs.

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