Question

evaluate the derivative of the following function.

$f(t)=\ln(\tan^{-1}3t^{2})$

$f(t)=\square$

Explanation:

Step1: Use the chain rule

The chain rule states that if \(y = f(u)\) and \(u = g(t)\), then \(\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}\). Let \(u = \tan^{- 1}(3t^{2})\), so \(y=\ln(u)\). First, find \(\frac{dy}{du}\) and \(\frac{du}{dt}\).
For \(y = \ln(u)\), \(\frac{dy}{du}=\frac{1}{u}\).
For \(u=\tan^{-1}(3t^{2})\), use the formula \(\frac{d}{dx}\tan^{-1}(x)=\frac{1}{1 + x^{2}}\). Let \(x = 3t^{2}\), then \(\frac{du}{dt}=\frac{6t}{1+(3t^{2})^{2}}=\frac{6t}{1 + 9t^{4}}\).

Step2: Substitute \(u\) and combine the derivatives

Since \(u=\tan^{-1}(3t^{2})\), \(\frac{dy}{dt}=\frac{1}{\tan^{-1}(3t^{2})}\cdot\frac{6t}{1 + 9t^{4}}\).

Answer:

\(\frac{6t}{(1 + 9t^{4})\tan^{-1}(3t^{2})}\)