Find the derivative of the function
g(x)=S(u^2-5)/(u^2+5)du
g'(x)=?
S=integration symbol
b=3x
a=2x
the 2nd FT of Calculus is just the chain rule in reverse.
If F(x) = ∫f(x) dx
∫[u,v] f(t) dt = F(u)-F(v)
so, taking derivatives,
F'(x) = dF/dv * dv/dx - dF/du * du/dx
= f(v) v' - f(u) u'
So, for this problem,
g(x) = ∫[2x,3x] f(u) du
g'(x) = f(3x)*3 - f(2x)*2
= 3((3x)^2-5)/((3x)^2+5) - 2((2x)^2-5)/((2x)^2+5)
and you can simplify that in a couple of ways. But that's just algebra...
To find the derivative of the function g(x) = ∫[a, b] (u^2-5)/(u^2+5) du, where a = 2x and b = 3x, we can use the Fundamental Theorem of Calculus.
First, let's find the antiderivative of the integrand (u^2-5)/(u^2+5) with respect to u. We can do this by using partial fraction decomposition.
(u^2-5)/(u^2+5) = (u^2-5)/(u^2+5)
To find the derivative of g(x), we'll evaluate g(x) at the upper limit b = 3x and then subtract the value of g(x) at the lower limit a = 2x.
g'(x) = [d/dx ∫[a, b] (u^2-5)/(u^2+5) du]
= [(d/dx) G(u)] |[a=2x, b=3x] (where G(u) is the antiderivative of (u^2-5)/(u^2+5))
Now, let's differentiate the antiderivative G(u) with respect to u.
(d/dx) G(u) = d/dx ∫(u^2-5)/(u^2+5) du
To differentiate G(u), we can use the chain rule by treating u as a function of x.
(d/dx) G(u) = (d/dx) G(u) * du/dx
Note that du/dx represents the derivative of u with respect to x.
(d/dx) G(u) * du/dx = (u^2-5)/(u^2+5) * du/dx
Now, we substitute u = 3x and u = 2x into the equation to evaluate the derivative at the limits.
g'(x) = [(u^2-5)/(u^2+5) * du/dx] |[2x, 3x]
Evaluating the expression at the upper limit (u = 3x):
[(3x^2-5)/(9x^2+5) * du/dx] |[2x, 3x]
Next, we evaluate the expression at the lower limit (u = 2x):
[(2x^2-5)/(4x^2+5) * du/dx] |[2x, 3x]
Simplifying this expression gives us the derivative of g(x), g'(x):
g'(x) = [(3x^2-5)/(9x^2+5)] - [(2x^2-5)/(4x^2+5)]