[Integrals]

Question

h(x)= -4 to sin(x) (cos(t^5)+t)dt

h'(x)=?

Answer 1

Not sure if you want to find the integral or the derivative. Please respond.
h'(x) generally stand for the derivative of function h(x).

Answer 2

Derivative.

Answer 3

By the definition of differentiation and integration, in general, if
I(x)=∫f(x)dx, then
I'(x)=f(x)

Can you take it from here?

Answer 4

Nope.

Answer 5

The question defines
h(x)=∫f(t)dt
where f(t)=cos(t^5)+t
and the limits of integration are from -4 to sin(x).

For experimentation, try f(t)=t;
h(x)=∫t;dt
=[t²/2]
to be evaluated between -4 and sin(x), which gives
h(x)=[sin²(x)/2 - (-4)²/2]
Now calculate h'(x) by the chain rule, namely for the first expression, differentiate with respect to sin(x), and then multiply by d(sin(x))/dx to get
h'(x)=2sin(x).cos(x)/2+0
=sin(x).cos(x)
You will notice that we have integrated t to get t²/2 and we differentiate sin²(x)/2 to get back sin(x).
What can you say about h'(x)?
If it is not clear, repeat the exercise using f(x)=cos(x), and see what you get for h'(x).