Find $\frac { d y } { d x }$ of $y = \sqrt { u }$ , $u = 9 - x ^ { 2 }$ .
A) $\frac { x } { \sqrt { 9 - x ^ { 2 } } }$
B) $\frac { 1 } { 2 \sqrt { 9 - x ^ { 2 } } }$
C) $\frac { - x } { \sqrt { 9 - x ^ { 2 } } }$
D) $- \frac { 1 } { 2 \sqrt { 9 - x ^ { 2 } } }$
E)none of these choices

Question

Quizplus · Accepted Answer

To find $\frac { d y } { d x }$ for $y = \sqrt { u }$ where $u = 9 - x ^ { 2 }$, we use the chain rule. First, find $\frac{dy}{du} = \frac{1}{2\sqrt{u}}$ and $\frac{du}{dx} = -2x$. Then, $\frac { d y } { d x } = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{9-x^2}} \cdot (-2x) = \frac{-x}{\sqrt{9-x^2}}$.

Find dydx\frac { d y } { d x }dxdy​ Of y=uy = \sqrt { u }y=u​

Find $\frac { d y } { d x }$ Of $y = \sqrt { u }$