Question

$\bold{Hey\ Brainiacs\ !}$ Find derivative $\rm \displaystyle y=\int_{0}^{ lnx} sine^t\ dt$ $\rm \orange{Quality\ answer\ needed}$

Answer 1

Given

$\rm \displaystyle y=\int_{0}^{ lnx} sine^t\ dt$

To Find

the derivative of y

Concept

Newton lebinitz formula is to be used which is

$\rm \dfrac{d \left(\displaystyle \int_{a}^{ b} f(t)dt \right)}{dx}\:=\: \dfrac{d(b)}{dx}\times f(b)\:-\:\dfrac{d(a)}{dx}\times f(a)$

Solution

$\rm \displaystyle y=\int_{0}^{ lnx} sine^t\ dt \\ \\ \\ \implies \:\dfrac{dy}{dx}\:=\:\dfrac{d(lnx)}{dx}\times sine^{lnx}\:-\:0 \\ \\ \\ \implies \: \dfrac{dy}{dx}\:=\: \dfrac{1}{x}\times sinx \:\:\:(\because e^{lnx}\:=\:x) \\ \\ \\ \implies \: \dfrac{dy}{dx}\:=\: \dfrac{sinx}{x}$

$\boxed{\dfrac{dy}{dx}\:=\:\dfrac{sinx}{x}}$

Answer 2

$\orange{\bigstar}$  Answer  $\green{\bigstar}$

$\rm y=\int_{0}^{lnx} sine^tdt$

Apply Newton Leibniz rule ,

$:\to \rm \dfrac{dy}{dx}=\dfrac{d}{dx}(\ln x)sin\ e^{\ln x}-\dfrac{d}{dx}(0)sin\ e^0\\\\ \boxed{\bf \because\ e^{ln\ x}=x\ }\\\\ :\to \rm \dfrac{dy}{dx}=\dfrac{1}{x}(sin\ x)-0\ \\\\ :\to \rm \blue{\bigstar}\ \; \dfrac{dy}{dx}=\dfrac{sin\ x}{x}\ \; \red{\bigstar}$