Question

differentiate the function.
f(x) = sin(3 ln(x))
f(x) =

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1. -/1 points

differentiate the function.
f(x) = ln(1/x^3)
f(x) =

Explanation:

Step1: Apply chain - rule for $y = \sin(3\ln(x))$

Let $u = 3\ln(x)$. Then $y=\sin(u)$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$. The derivative of $y = \sin(u)$ with respect to $u$ is $\frac{dy}{du}=\cos(u)$. The derivative of $u = 3\ln(x)$ with respect to $x$ is $\frac{du}{dx}=\frac{3}{x}$.

Step2: Substitute $u$ back and find $\frac{dy}{dx}$

Substitute $u = 3\ln(x)$ into $\frac{dy}{du}\cdot\frac{du}{dx}$. We get $\frac{dy}{dx}=\cos(3\ln(x))\cdot\frac{3}{x}=\frac{3\cos(3\ln(x))}{x}$.

Step3: Simplify $f(x)=\ln(\frac{1}{x^{3}})$ using logarithm property

We know that $\ln(\frac{1}{x^{3}})=\ln(1)-\ln(x^{3})$. Since $\ln(1) = 0$ and $\ln(x^{3})=3\ln(x)$ by the power - rule of logarithms, then $f(x)=- 3\ln(x)$.

Step4: Differentiate $f(x)=-3\ln(x)$

The derivative of $\ln(x)$ is $\frac{1}{x}$. Using the constant - multiple rule of differentiation, if $y=-3\ln(x)$, then $y^\prime=-3\cdot\frac{1}{x}=-\frac{3}{x}$.

Answer:

For $f(x)=\sin(3\ln(x))$, $f^\prime(x)=\frac{3\cos(3\ln(x))}{x}$; for $f(x)=\ln(\frac{1}{x^{3}})$, $f^\prime(x)=-\frac{3}{x}$