For $y = u \mathcal { \sqrt { u + 1 } }$ and $u = 2 x ^ { 2 } + 3$ , find $\frac { d y } { d x }$ .
A) $\left( 2 x ^ { 2 } + 4 \right) ^ { - 1 / 2 } \left( 12 x ^ { 3 } + 22 x \right)$
B) $\left( 2 x ^ { 2 } + 3 \right) ^ { - 1 / 2 } \left( 12 x ^ { 3 } + 24 x \right)$
C) $\left( 2 x ^ { 2 } + 4 \right) ^ { - 1 / 2 } ( 4 x )$
D) $\left( 2 x ^ { 2 } + 3 \right) ^ { - 1 / 2 } ( 4 x )$

Question

Quizplus · Accepted Answer

To find $\frac{dy}{dx}$, we use the chain rule. First, differentiate $y$ with respect to $u$, getting $y_u = \sqrt{u + 1} + \frac{u}{2\sqrt{u + 1}}$. Then, differentiate $u$ with respect to $x$, getting $u_x = 4x$. The product of these derivatives, multiplied by the derivative of $u$ with respect to $x$, gives the correct answer. Simplifying the expression leads to the formula in option A.

For y=uu+1y = u \mathcal { \sqrt { u + 1 } }y=uu+1​

For $y = u \mathcal { \sqrt { u + 1 } }$