Question

find $\frac{dy}{dx}$. $y = \tan^{-1}(ln(4x^{3}))$ $\frac{dy}{dx}=square$ (use parentheses to clearly denote the argument of

Explanation:

Step1: Apply chain - rule

Let $u = \ln(4x^{3})$, then $y=\tan^{- 1}(u)$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$. The derivative of $y = \tan^{-1}(u)$ with respect to $u$ is $\frac{dy}{du}=\frac{1}{1 + u^{2}}$.

Step2: Find $\frac{du}{dx}$

Since $u=\ln(4x^{3})$, and using the property $\ln(ab)=\ln(a)+\ln(b)$, we have $u=\ln(4)+3\ln(x)$. The derivative of $\ln(4)$ (a constant) with respect to $x$ is $0$, and the derivative of $3\ln(x)$ with respect to $x$ is $\frac{3}{x}$. So, $\frac{du}{dx}=\frac{3}{x}$.

Step3: Substitute $u$ and calculate $\frac{dy}{dx}$

Substitute $u = \ln(4x^{3})$ into $\frac{dy}{du}$ and multiply by $\frac{du}{dx}$. We get $\frac{dy}{dx}=\frac{1}{1+(\ln(4x^{3}))^{2}}\cdot\frac{3}{x}=\frac{3}{x(1 + (\ln(4x^{3}))^{2})}$.

Answer:

$\frac{3}{x(1 + (\ln(4x^{3}))^{2})}$