Question
[Integrals]
h(x)= -4 to sin(x) (cos(t^5)+t)dt
h'(x)=?
h(x)= -4 to sin(x) (cos(t^5)+t)dt
h'(x)=?
Answers
MathMate
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).
h'(x) generally stand for the derivative of function h(x).
Z32
Derivative.
MathMate
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?
I(x)=∫f(x)dx, then
I'(x)=f(x)
Can you take it from here?
Z32
Nope.
MathMate
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).
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).