Question

write the function below in the form ( y = f(u) ) and ( u = g(x) ). then find ( \frac{dy}{dx} ) as a function of ( x ).
( y=-8(sin x)^{-7} )
write the function in the form ( y = f(u) ) and ( u = g(x) ). choose the correct answer below.
a. ( y=-8 u^{-7} ) and ( u=sin x )
b. ( y = u ) and ( u=sin x )
c. ( y=sin u ) and ( u = x )
d. ( y=sin u ) and ( u=-8 x^{-7} )

Explanation:

Step1: Identify \(y = f(u)\) and \(u = g(x)\)

Given \(y=-8(\sin x)^{-7}\), if we let \(u = \sin x\), then \(y=-8u^{-7}\). So \(y = f(u)=-8u^{-7}\) and \(u = g(x)=\sin x\), which is option A.

Step2: Use the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\)

First, find \(\frac{dy}{du}\):
If \(y=-8u^{-7}\), then by the power rule \(\frac{dy}{du}=-8\times(-7)u^{-8}=56u^{-8}\) (since \(\frac{d}{du}(au^{n})=an u^{n - 1}\), here \(a=-8\) and \(n=-7\)).
Second, find \(\frac{du}{dx}\):
If \(u = \sin x\), then \(\frac{du}{dx}=\cos x\) (because \(\frac{d}{dx}(\sin x)=\cos x\)).

Step3: Substitute \(u=\sin x\) into \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\)

\(\frac{dy}{dx}=56u^{-8}\cdot\cos x\). Substituting \(u = \sin x\) gives \(\frac{dy}{dx}=\frac{56\cos x}{(\sin x)^{8}}\)

Answer:

A. \(y = - 8u^{-7}\) and \(u=\sin x\); \(\frac{dy}{dx}=\frac{56\cos x}{(\sin x)^{8}}\)