Answer:
[tex]g'(x) = \dfrac{5 *(25x^2-1)}{(25x^2+1)}- \dfrac{4 * (16x^2-1)}{(16x^2+1)}[/tex]
Step-by-step explanation:
Given that:
[tex]g(x) = \int^{5x}_{4x} \dfrac{u^2-1}{u^2+1} \ \ du[/tex]
Then: the derivative of the function is as follows:
[tex]g'(x) = \dfrac{(5x)^2-1}{(5x)^2+1} \ \ \dfrac{d}{dx}(5x) -\dfrac{(4x)^2-1}{(4x)^2+1} \ \ \dfrac{d}{dx}(4x)[/tex]
[tex]g'(x) = \dfrac{5 *(25x^2-1)}{(25x^2+1)}- \dfrac{4 * (16x^2-1)}{(16x^2+1)}[/tex]
