Find the derivative of the following function using the appropriate form of the Fundamental Theorem of Calculus.
F(x)= (s^2)/(1+3s^4)ds from sqrtx to 1
F'(x)=?
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To find the derivative of the function F(x), we need to apply the appropriate form of the Fundamental Theorem of Calculus.
First, let's rewrite the integral in the function as follows:
F(x) = ∫[(s^2)/(1+3s^4)]ds from sqrt(x) to 1
To find F'(x), we'll use the second part of the Fundamental Theorem of Calculus, which states that if a function F(x) is defined as the integral of another function f(t) from a constant a to a variable x, then F'(x) is equal to f(x).
In this case, we have F(x) defined as the integral of (s^2)/(1+3s^4) ds from sqrt(x) to 1. So, we need to find the derivative of the integrand (s^2)/(1+3s^4) with respect to s.
Let's denote the integrand as g(s):
g(s) = (s^2)/(1+3s^4)
To find the derivative of g(s) with respect to s, we'll use the quotient rule, which states that for two functions u(s) and v(s), the derivative of their quotient is given by:
(u'v - uv') / (v^2)
In our case, u(s) = s^2 and v(s) = 1+3s^4. Therefore, we have:
u'(s) = 2s
v'(s) = 12s^3
Plugging these values into the quotient rule, we get:
g'(s) = [(2s)(1+3s^4) - (s^2)(12s^3)] / (1+3s^4)^2
Now, to find F'(x), we substitute back the appropriate variables:
F'(x) = g'(1) - g'(sqrt(x))
Substitute s=1 into g'(s):
g'(1) = [(2(1))(1+3(1)^4) - (1^2)(12(1)^3)] / (1+3(1)^4)^2
Next, substitute s=sqrt(x) into g'(s):
g'(sqrt(x)) = [(2(sqrt(x)))(1+3(sqrt(x))^4) - (sqrt(x))^2(12(sqrt(x))^3)] / (1+3(sqrt(x))^4)^2
Finally, we have:
F'(x) = g'(1) - g'(sqrt(x)) = [(2(1))(1+3(1)^4) - (1^2)(12(1)^3)] / (1+3(1)^4)^2 - [(2(sqrt(x)))(1+3(sqrt(x))^4) - (sqrt(x))^2(12(sqrt(x))^3)] / (1+3(sqrt(x))^4)^2
This is the expression for F'(x), the derivative of the function F(x), using the appropriate form of the Fundamental Theorem of Calculus.