Find the derivative.
-$y = \int _ { 0 } ^ { \sqrt { x } } 9 t \cos \left( t ^ { 8 } \right) d t$
A) $9 \cos \left( x ^ { 4 } \right) - 9 \sqrt { x } \sin \left( x ^ { 4 } \right)$
B) $8 \sqrt { x } \cos \left( x ^ { 4 } \right)$
C) $9 \sqrt { x } \cos \left( x ^ { 4 } \right)$
D) $\frac { 9 } { 2 } \cos \left( x ^ { 4 } \right)$

Question

Quizplus · Accepted Answer

The derivative of the integral with respect to $x$ can be found using the Fundamental Theorem of Calculus Part 1, which states that if $F(x) = \int_{a}^{g(x)} f(t) dt$, then $F'(x) = f(g(x)) \cdot g'(x)$. Here, $f(t) = 9t\cos(t^8)$ and $g(x) = \sqrt{x}$, so $g'(x) = \frac{1}{2\sqrt{x}}$. Substituting $g(x)$ into $f(t)$ gives $9\sqrt{x}\cos((\sqrt{x})^8) = 9\sqrt{x}\cos(x^4)$, and multiplying by $g'(x)$ gives $\frac{9}{2}\cos(x^4)$.

Find the Derivative y=∫0x9tcos⁡(t8)dty = \int _ { 0 } ^ { \sqrt { x } } 9 t \cos \left( t ^ { 8 } \right) d ty=∫0x​​9tcos(t8)dt

Find the Derivative $y = \int _ { 0 } ^ { \sqrt { x } } 9 t \cos \left( t ^ { 8 } \right) d t$