evaluate the derivative of the following function.\n\n$f(t)=\\ln(\\tan^{-1}3t^{2})$\n\n$f(t)=\\square$
evaluate the derivative of the following function.\n\n$f(t)=\\ln(\\tan^{-1}3t^{2})$\n\n$f(t)=\\square$
Answer
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})})