Respuesta :

[tex]\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}\\\\ -----------------------------\\\\[/tex]

[tex]\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}[/tex]
apologiabiology
2nd theorem of calculus says
[tex] \frac{dy}{dx} \int\limits^z_a {f(t)} \, dt =f(x)(z) \frac{dy}{dx}[/tex] when a is a constant

so we are given
a=x^2
and we want to do
[tex] \frac{dy}{dx} \int\limits^x_a {\frac{t}{t+1}} \, dt =\frac{x^2}{x^2+1}(x^2)\frac{dy}{dx}=[/tex][tex]\frac{x^2}{x^2+1}2x=\frac{2x^3}{x^2+1}[/tex]

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