Write the function in the form y = f(u) and u = g(x). Then find dy/dx as a function of x.
-$$y = \tan \left( \pi - \frac { 4 } { x } \right)$$
A) $y = \tan u ; u = \pi - \frac { 4 } { x } ; \frac { d y } { d x } = \sec ^ { 2 } \left( \frac { 4 } { x ^ { 2 } } \right)$
B) $y = \tan u ; u = \pi - \frac { 4 } { x } ; \frac { d y } { d x } = \frac { 4 } { x ^ { 2 } } \sec ^ { 2 } \left( \pi - \frac { 4 } { x } \right)$
C) $y = \tan u ; u = \pi - \frac { 4 } { x } ; \frac { d y } { d x } = \sec ^ { 2 } \left( \pi - \frac { 4 } { x } \right)$
D) $y = \tan u ; u = \pi - \frac { 4 } { x } ; \frac { d y } { d x } = \frac { 4 } { x ^ { 2 } } \sec \left( \pi - \frac { 4 } { x } \right) \tan \left( \pi - \frac { 4 } { x } \right)$

Question

Quizplus · Accepted Answer

We can rewrite the function as $y = 	an(\pi - \frac{4}{x})$. Let $u = \pi - \frac{4}{x}$, then $y = 	an u$. Using the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

We know that $\frac{dy}{du} = \sec^2 u$, and $\frac{du}{dx} = \frac{4}{x^2}$. Substituting back in for u, we get $\frac{dy}{dx} = \frac{4}{x^2}\sec^2(\pi - \frac{4}{x})$. Using the identity $\sec^2 	heta = 1 + 	an^2 	heta$, we can simplify to get $\frac{dy}{dx} = \frac{4}{x^2}(1 + 	an^2(\frac{4}{x}))$. Finally, we can substitute $	an^2 	heta = \sec^2 	heta - 1$ to get $\frac{dy}{dx} = \frac{4}{x^2} \sec^2(\frac{4}{x})$. Therefore, the best choice is B.

Write the Function in the Form Y = F(u) and U