Let $f ( x ) = \int _ { x ^ { 2 } } ^ { 3 } \sqrt [ 3 ] { 4 - t ^ { 3 } } d t . \text { Find } f ^ { \prime } ( x )$
A)0
B) $- 2 x \sqrt [ 3 ] { 4 - x ^ { 6 } }$
C) $2 x \sqrt [ 3 ] { 4 - x ^ { 6 } }$
D) $\sqrt [ 3 ] { 4 - x ^ { 3 } }$
E) $- \sqrt [ 3 ] { 4 - x ^ { 6 } }$
F)
$- 2 x \sqrt [ 3 ] { 4 - x ^ { 3 } }$
G) $\sqrt [ 3 ] { 4 - x ^ { 6 } }$
H) $- \sqrt [ 3 ] { 4 - x ^ { 3 } }$

Question

Let $f ( x ) = \int _ { x ^ { 2 } } ^ { 3 } \sqrt [ 3 ] { 4 - t ^ { 3 } } d t . \text { Find } f ^ { \prime } ( x )$ 
A)0
B) $- 2 x \sqrt [ 3 ] { 4 - x ^ { 6 } }$
C) $2 x \sqrt [ 3 ] { 4 - x ^ { 6 } }$
D) $\sqrt [ 3 ] { 4 - x ^ { 3 } }$
E) $- \sqrt [ 3 ] { 4 - x ^ { 6 } }$  
F)
$- 2 x \sqrt [ 3 ] { 4 - x ^ { 3 } }$  
G) $\sqrt [ 3 ] { 4 - x ^ { 6 } }$  
H) $- \sqrt [ 3 ] { 4 - x ^ { 3 } }$

Quizplus · Accepted Answer

To find $f'(x)$, we use the Fundamental Theorem of Calculus part 1 and the chain rule. The derivative of the integral with respect to $x$ involves evaluating the integrand at the upper limit of integration (which is a function of $x$) and then multiplying by the derivative of this upper limit, minus evaluating the integrand at the lower limit of integration (also a function of $x$) and then multiplying by the derivative of this lower limit. Since the upper limit is a constant (3), its derivative is 0, and we only need to consider the lower limit $x^2$. Thus, $f'(x) = -\sqrt[3]{4 - (x^2)^3} \cdot \frac{d}{dx}(x^2) = -2x\sqrt[3]{4 - x^6}$.

Let f(x)=∫x234−t33dt. Find f′(x)f ( x ) = \int _ { x ^ { 2 } } ^ { 3 } \sqrt [ 3 ] { 4 - t ^ { 3 } } d t . \text { Find } f ^ { \prime } ( x )f(x)=∫x23​34−t3​dt. Find f′(x)

Let $f ( x ) = \int _ { x ^ { 2 } } ^ { 3 } \sqrt [ 3 ] { 4 - t ^ { 3 } } d t . \text { Find } f ^ { \prime } ( x )$