$$\text { Find the derivative of } y \text { with respect to } x , t , \text { or } \theta \text {, as appropriate. }$$
-$$y = e ^ { \left( 2 \sqrt { x } + x ^ { 5 } \right) }$$
A) $\left( \frac { 1 } { \sqrt { x } } + 5 x ^ { 4 } \right) e ^ { \left( 2 \sqrt { x } + x ^ { 5 } \right) }$
B) $\left( 2 \sqrt { x } + 5 x ^ { 4 } \right) \ln \left( 2 \sqrt { x + } x ^ { 5 } \right)$
C) $\left. 2 \sqrt { x + 5 } x ^ { 4 } \right) e ^ { \left( 2 \sqrt { x } + x ^ { 5 } \right) }$
D) $e ^ { \left( 1 \sqrt { x } + 5 x ^ { 4 } \right) }$

Question

Quizplus · Accepted Answer

Using the chain rule, we have:
$$\frac{dy}{dx} = e^{2\sqrt{x}+x^5}\cdot\left(\frac{d}{dx}\left(2\sqrt{x}+ x^5\right)\right)$$
$$= e^{2\sqrt{x}+x^5}\cdot\left(2\cdot\frac{1}{2\sqrt{x}}+5x^4\right)$$
$$= e^{2\sqrt{x}+x^5}\cdot\left(\frac{1}{\sqrt{x}}+5x^4\right)$$
Therefore, the answer is (A).

Find the derivative of y with respect to x,t, or θ, as appropriate. \text { Find the derivative of } y \text { with respect to } x , t , \text { or } \theta \text {, as appropriate. } Find the derivative of y with respect to x,t, or θ, as appropriate.

$\text { Find the derivative of } y \text { with respect to } x , t , \text { or } \theta \text {, as appropriate. }$