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find the derivative of $f(x)=\ln(x^4)(\sqrt{5x-3})$

I just need help getting to the answer. The first answer I got was $f(x)=(x^4)(2.5)+(5x-3)^{1/2}(4x^3)$.

Emilio Novati
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alex
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1 Answers1

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We will first use the product rule:

$$\frac{d}{dx} \ln(x^4)\sqrt{5x-3} = \ln(x^4)\cdot\frac{d}{dx}\sqrt{5x-3} + \sqrt{5x-3}\cdot\frac{d}{dx}\ln(x^4)$$

Then we use the chain rule on the first part. $u=5x-3$.

\begin{align}\ln(x^4)\frac{d}{dx}\sqrt{5x-3} & =\ln(x^4)\frac{d}{du}\sqrt{u} \frac{d}{dx}5x-3 \\ &= 5\ln(x^4) 0.5 u^{-\frac{1}{2}} \\ &=2.5 \ln(x^4)(5x-3)^{-\frac{1}{2}}\end{align}

Then we use the chain rule on the second part. $v=x^4$.

\begin{align}\sqrt{5x-3}\cdot\frac{d}{dx}\ln(x^4) &=\sqrt{5x-3}\cdot\frac{d}{dv}\ln(v)\cdot\frac{d}{dx}x^4 \\ &= \sqrt{5x-3}\cdot\frac{1}{v}\cdot4x^3 \\ &= 4x^3\cdot\sqrt{5x-3}\cdot\frac{1}{x^4} \\ &= 4 \frac{1}{x}\cdot\sqrt{5x-3} \end{align}

So the derivative is $$2.5 \ln(x^4)(5x-3)^{-\frac{1}{2}} + 4 \frac{1}{x}\sqrt{5x-3}$$

EDIT: You could also simplify $\ln(x^4) = 4 \ln(x)$, that would have saved one chain rule.

EDIT 2: I think you went wrong on the product rule. Also (I suppose you know this but I will say it): The derivative of $\ln(x)$ is $\frac{1}{x}$. This can be proven elegantly by the chain rule if you know that the derivative of $e^x$ is $e^x$.

wythagoras
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