Find the derivative.
-$$y = \int _ { 5 ^ { x } } ^ { 1 } \sqrt [ 4 ] { t } d t$$
A) $- 5 ( 1 / 4 ) x$
B) $- \frac { 5 ^ { ( 5 / 4 ) \times } 25 \ln 5 } { 16 }$
C) $- 5 ( 3 / 4 ) x \ln 5$
D) $- 5 ( 5 / 4 ) \mathrm { x } \ln 5$

Question

Quizplus · Accepted Answer

Using the Fundamental Theorem of Calculus and the Chain Rule, we have:

$$y' = \frac{d}{dx} \int _ { 5 ^ { x } } ^ { 1 } \sqrt [ 4 ] { t } d t = \frac{d}{dx} \left[-\int _1^{5^x} t^{1/4} dt\right] = -\frac{d}{dx}\left(\frac{5^{x+1}-1}{\frac{1}{4}}\right)$$

$$= -\frac{5^{x+1}\ln 5}{4} = -\frac{5^{(5/4)x+1} \ln 5}{4} = -5^{(5/4)x} \cdot \frac{5\ln 5}{4} = -\frac{5^{(5/4)x+2}\ln 5}{16}.$$

Therefore, the answer is D.

Find the Derivative y=∫5x1t4dty = \int _ { 5 ^ { x } } ^ { 1 } \sqrt [ 4 ] { t } d ty=∫5x1​4t​dt

Find the Derivative $y = \int _ { 5 ^ { x } } ^ { 1 } \sqrt [ 4 ] { t } d t$