Find the derivative of the function $y = \ln \sqrt { x ^ { 2 } - 16 }$ .
A) $\frac { d y } { d x } = \frac { 1 } { 2 x - 16 }$
B) $\frac { d y } { d x } = \frac { x } { 2 x - 16 }$
C) $\frac { d y } { d x } = \frac { 1 } { x ^ { 2 } - 16 }$
D) $\frac { d y } { d x } = \frac { x } { x ^ { 2 } - 16 }$
E) $\frac { d y } { d x } = \frac { 2 x } { \sqrt { x ^ { 2 } - 16 } }$

Question

Quizplus · Accepted Answer

The derivative of $y = \ln \sqrt{x^2 - 16}$ is found using the chain rule: $\frac{d}{dx}(\ln u) = \frac{1}{u} \cdot \frac{du}{dx}$. Here, $u = \sqrt{x^2 - 16}$, so $\frac{du}{dx} = \frac{x}{\sqrt{x^2 - 16}}$. Thus, $\frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 16}} \cdot \frac{x}{\sqrt{x^2 - 16}} = \frac{x}{x^2 - 16}$.

Find the Derivative of the Function y=ln⁡x2−16y = \ln \sqrt { x ^ { 2 } - 16 }y=lnx2−16​

Find the Derivative of the Function $y = \ln \sqrt { x ^ { 2 } - 16 }$