Question

find $\frac{dy}{dx}$. $y = \tan^{-1}(ln(3x^{5}))$ $\frac{dy}{dx}=square$ (use parentheses to clearly denote the argument of ea)

Explanation:

Step1: Apply chain - rule

Let $u = \ln(3x^{5})$, 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 $u=\ln(3x^{5})=\ln(3)+5\ln(x)$ (using the property $\ln(ab)=\ln(a)+\ln(b)$ and $\ln(a^{b})=b\ln(a)$), the derivative of $u$ with respect to $x$ is $\frac{du}{dx}=\frac{5}{x}$.

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

Substitute $u = \ln(3x^{5})$ into $\frac{dy}{du}$ and multiply by $\frac{du}{dx}$. We get $\frac{dy}{dx}=\frac{1}{1+(\ln(3x^{5}))^{2}}\cdot\frac{5}{x}=\frac{5}{x(1 + (\ln(3x^{5}))^{2})}$.

Answer:

$\frac{5}{x(1 + (\ln(3x^{5}))^{2})}$