Differentiate $f ( w ) = \sqrt { w ^ { 2 } + 8 }$ .
A) $\frac { 8 w } { \sqrt { w ^ { 2 } + 8 } }$
B) $\sqrt { 2 w }$
C) $\frac { w } { \sqrt { w ^ { 2 } + 8 } }$
D) $\frac { 1 } { 2 \sqrt { w ^ { 2 } + 8 } }$

Question

Quizplus · Accepted Answer

To differentiate $f(w) = \sqrt{w^2 + 8}$, we use the chain rule. Let $u = w^2 + 8$, then $f(w) = \sqrt{u}$. The derivative of $f$ with respect to $w$ is $f'(w) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dw}$. Since $u = w^2 + 8$, $\frac{du}{dw} = 2w$. Thus, $f'(w) = \frac{1}{2\sqrt{w^2 + 8}} \cdot 2w = \frac{w}{\sqrt{w^2 + 8}}$.

Differentiate f(w)=w2+8f ( w ) = \sqrt { w ^ { 2 } + 8 }f(w)=w2+8​

Differentiate $f ( w ) = \sqrt { w ^ { 2 } + 8 }$