[Integrals]
h(x)= -4 to sin(x) (cos(t^5)+t)dt
h'(x)=?
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).
Derivative.
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?
Nope.
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).
To find the derivative of the function h(x), we will use the Fundamental Theorem of Calculus and apply the chain rule.
The Fundamental Theorem of Calculus states that if we have a function h(x) defined as the integral of another function F(x) with respect to t, then the derivative of h(x) with respect to x is equal to F(sin(x)) times the derivative of the function inside the integral.
In this case, h(x) is defined as the integral from -4 to sin(x) of the function (cos(t^5) + t) dt. Let's call the function inside the integral F(t), so F(t) = cos(t^5) + t.
To find h'(x), we need to calculate F(sin(x)) times the derivative of sin(x) with respect to x and evaluate it for x. Let's break it down step by step:
1. Calculate F(sin(x)) by substituting sin(x) into F(t):
F(sin(x)) = cos((sin(x))^5) + sin(x)
2. Calculate the derivative of sin(x) with respect to x:
sin'(x) = cos(x)
3. Multiply F(sin(x)) by sin'(x):
F(sin(x)) * sin'(x) = (cos((sin(x))^5) + sin(x)) * cos(x)
Finally, we can say that h'(x) is equal to F(sin(x)) * sin'(x):
h'(x) = (cos((sin(x))^5) + sin(x)) * cos(x)
So, the derivative of h(x) with respect to x is (cos((sin(x))^5) + sin(x)) * cos(x).