Question

Calculate the derivative of


`\int\limits^a_ 0{(tdt)/(t+1) \,`

a =
`x^(2)`

Answer 1

2nd theorem of calculus says

`(dy)/(dx) \int\limits^z_a {f(t)} \, dt =f(x)(z) (dy)/(dx)` when a is a constant

so we are given
a=x^2
and we want to do

`(dy)/(dx) \int\limits^x_a {(t)/(t+1)} \, dt =(x^2)/(x^2+1)(x^2)(dy)/(dx)=`
`(x^2)/(x^2+1)2x=(2x^3)/(x^2+1)`

Answer 2

`\bf \cfrac{dy}{dx}=\cfrac{dy}{du}\cdot \cfrac{du}{dx}\\ \left. \qquad \right.\uparrow \\ \textit{2nd fundamental theorem of calculus}\\\\ -----------------------------\\\\ \displaystyle \cfrac{dy}{dt}\left( \int\limits_(0)^(x^2)\ \cfrac{t}{t+1}\cdot dt \right)\implies \cfrac{dy}{du}\cdot \cfrac{du}{dx}\\\\ -----------------------------\\\\`


`\bf u=x^2\qquad \cfrac{du}{dx}=2x\\\\ -----------------------------\\\\ \displaystyle \cfrac{dy}{dt}\left( \int\limits_(0)^(u)\ \cfrac{t}{t+1}\cdot dt \right)\implies \cfrac{u}{u+1}\cdot \cfrac{du}{dx}\implies \cfrac{u}{u+1}\cdot 2x \\\\\\ u=x^2\qquad thus\implies \cfrac{x^2}{x^2+1}\cdot 2x\implies \cfrac{2x^3}{x^2+1}`