Question

question 5 (1 point)
if ( g(x)=int_{x}^{7} cos left(t^{7}
ight) d t ), then ( g^{prime}(x)= )
( cos left(x^{7}
ight) )
( cos left(7^{7}
ight)-cos left(x^{7}
ight) )
( -sin left(x^{7}
ight) )
( -sin left(7^{7}
ight)+sin left(x^{7}
ight) )
( -cos left(x^{7}
ight) )
view hint for question 5

Explanation:

Step1: Recall Leibniz Rule

For $g(x)=\int_{a(x)}^{b(x)} f(t)dt$, $g'(x)=f(b(x))b'(x)-f(a(x))a'(x)$

Step2: Identify limits and function

Here, $b(x)=7$, $a(x)=x$, $f(t)=\cos(t^7)$

Step3: Compute derivatives of limits

$b'(x)=0$, $a'(x)=1$

Step4: Substitute into formula

$g'(x)=\cos(7^7)\cdot0 - \cos(x^7)\cdot1 = -\cos(x^7)$

Answer:

$-\cos(x^7)$