Question

differentiate the function.
g(x) = ln(xe^(-7x))
g(x) =

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differentiate the function.
h(x) = ln(x + sqrt(x^2 - 1))
h(x) =

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Explanation:

Step1: Simplify $g(x)$ using log - rules

Use $\ln(ab)=\ln a+\ln b$. So $g(x)=\ln(xe^{-7x})=\ln x+\ln(e^{-7x})=\ln x - 7x$.

Step2: Differentiate term - by - term

The derivative of $\ln x$ is $\frac{1}{x}$ and the derivative of $-7x$ is $-7$. So $g^{\prime}(x)=\frac{1}{x}-7$.

Step3: Simplify $h(x)$ using the chain - rule

Let $u = x+\sqrt{x^{2}-1}$. Then $h(x)=\ln(u)$. The derivative of $\ln(u)$ with respect to $u$ is $\frac{1}{u}$.

Step4: Differentiate $u$ with respect to $x$

The derivative of $x$ is $1$, and the derivative of $\sqrt{x^{2}-1}=(x^{2}-1)^{\frac{1}{2}}$ using the chain - rule. Let $v = x^{2}-1$, then $\frac{d}{dx}(x^{2}-1)^{\frac{1}{2}}=\frac{1}{2}(x^{2}-1)^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{x^{2}-1}}$. So $\frac{du}{dx}=1 + \frac{x}{\sqrt{x^{2}-1}}=\frac{\sqrt{x^{2}-1}+x}{\sqrt{x^{2}-1}}$.

Step5: Use the chain - rule for $h(x)$

$h^{\prime}(x)=\frac{1}{u}\cdot\frac{du}{dx}$. Substituting $u = x+\sqrt{x^{2}-1}$ and $\frac{du}{dx}=\frac{\sqrt{x^{2}-1}+x}{\sqrt{x^{2}-1}}$, we get $h^{\prime}(x)=\frac{1}{x + \sqrt{x^{2}-1}}\cdot\frac{\sqrt{x^{2}-1}+x}{\sqrt{x^{2}-1}}=\frac{1}{\sqrt{x^{2}-1}}$.

Answer:

$g^{\prime}(x)=\frac{1}{x}-7$
$h^{\prime}(x)=\frac{1}{\sqrt{x^{2}-1}}$