Find the derivative: frac{d}{dx} { e^{-ax^2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To find the derivative of the product of two functions,we use the product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.Let $u(x) = e^{-a

Vedclass · Accepted Answer

$e^{-ax^2}(\cot x - 2ax \log \sin x)$

Vedclass · Answer

(C) To find the derivative of the product of two functions,we use the product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.Let $u(x) = e^{-ax^2}$ and $v(x) = \log(\sin x)$.First,find the derivatives of $u(x)$ and $v(x)$:$u'(x) = e^{-ax^2} \cdot \frac{d}{dx}(-ax^2) = e^{-ax^2} \cdot (-2ax) = -2ax e^{-ax^2}$.$v'(x) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.Now,apply the product rule:$\frac{d}{dx} \{ e^{-ax^2} \log(\sin x) \} = (-2ax e^{-ax^2}) \cdot \log(\sin x) + e^{-ax^2} \cdot \cot x$.Factor out $e^{-ax^2}$:$= e^{-ax^2} [\cot x - 2ax \log(\sin x)]$.Thus,the correct option is $C$.

Find the derivative: $\frac{d}{dx} \{ e^{-ax^2} \log(\sin x) \}$

Similar Questions

Find the derivative of $f$ given by $f(x) = \sin^{-1} x$ assuming it exists.

If $y=\log \left[\tan \sqrt{\frac{2^x-1}{2^x+1}}\right], x>0$,then $\left(\frac{d y}{d x}\right)_{x=1}=$

Differentiate the following function with respect to $x$:
$\sqrt{3x+2} + \frac{1}{\sqrt{2x^2+4}}$

If the slope of the tangent drawn at any point $(x, y)$ on the curve $y=f(x)$ is $(6x^2+10x-9)$ and $f(2)=0$,then $f(-2)=$

Find the derivative of the following function (it is to be understood that $a, b, c,$ and $d$ are fixed non-zero constants): $\frac{a x+b}{c x+d}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module