Question

Help finding derivative

Answer 1

`\bf \displaystyle \cfrac{d}{dx}\left[ \int\limits_(√(x))^{(\pi )/(2)}~[cos(t^2)]dt \right]\implies \cfrac{d}{dx}\left[ -\int\limits_{(\pi )/(2)}^(√(x))~[cos(t^2)]dt \right]\\\\\\ \cfrac{d}{dx}\left[ -\int\limits_{(\pi )/(2)}^{x^{(1)/(2)}}~[cos(t^2)]dt \right]\\\\ -------------------------------\\\\ u=x^{(1)/(2)}\implies \cfrac{du}{dx}=\cfrac{1}{2√(x)}\\\\\\ \textit{and now, let's use the \underline{2nd fundamental theorem of calculus}}\\\\ -------------------------------\\\\`


`\bf \displaystyle \cfrac{d}{dx}\left[ -\int\limits_{(\pi )/(2)}^(u)~[cos(t^2)]dt \right]\impliedby \cfrac{df}{dx}=\cfrac{df}{du}\cdot \cfrac{du}{dx} \\\\\\ \cfrac{df}{dx}=-cos(u^2)\cdot \cfrac{du}{dx}\implies \cfrac{df}{dx}=-cos(u^2)\cdot \cfrac{1}{2√(x)} \\\\\\ \cfrac{df}{dx}=-cos[(√(x))^2]\cdot \cfrac{1}{2√(x)}\implies \cfrac{df}{dx}=-\cfrac{cos(x)}{2√(x)}`