The value of frac{d}{dx} tan^{-1} left[ frac{ | vedclass

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

Question

(B) Let $y = 	an^{-1} \left[ \frac{3a^2x - x^3}{a(a^2 - 3x^2)} ight]$.Substitute $x = a 	an 	heta$,which implies $	heta = 	an^{-1}(\frac{x}{a})

Vedclass · Accepted Answer

$\frac{3}{a}$

Vedclass · Answer

(B) Let $y = 	an^{-1} \left[ \frac{3a^2x - x^3}{a(a^2 - 3x^2)} ight]$.Substitute $x = a 	an 	heta$,which implies $	heta = 	an^{-1}(\frac{x}{a})$.Then,$y = 	an^{-1} \left[ \frac{3a^3 	an 	heta - a^3 	an^3 	heta}{a(a^2 - 3a^2 	an^2 	heta)} ight] = 	an^{-1} \left[ \frac{a^3(3 	an 	heta - 	an^3 	heta)}{a^3(1 - 3 	an^2 	heta)} ight]$.Using the trigonometric identity $	an(3	heta) = \frac{3 	an 	heta - 	an^3 	heta}{1 - 3 	an^2 	heta}$,we get $y = 	an^{-1}(	an 3	heta) = 3	heta$.Since $	heta = 	an^{-1}(\frac{x}{a})$,we have $y = 3 	an^{-1}(\frac{x}{a})$.Differentiating with respect to $x$:$\frac{dy}{dx} = 3 \cdot \frac{1}{1 + (x/a)^2} \cdot \frac{1}{a} = 3 \cdot \frac{a^2}{a^2 + x^2} \cdot \frac{1}{a} = \frac{3a}{a^2 + x^2}$.At $x = 0$,$\frac{dy}{dx} = \frac{3a}{a^2 + 0} = \frac{3}{a}$.

The value of $\frac{d}{dx} \tan^{-1} \left[ \frac{3a^2x - x^3}{a(a^2 - 3x^2)} \right]$ at $x = 0$ is

Similar Questions

The numerical value of $\tan \left(2 \tan ^{-1}\left(\frac{1}{5}\right)+\frac{\pi}{4}\right)$ is:

If $f(x) = \operatorname{Sec}^{-1}\left(\frac{1}{2x^2-1}\right)$ and $g(x) = \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$,then the derivative of $f(x)$ with respect to $g(x)$ is

If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$,then $x(y+z)+y(z+x)+z(x+y)$ equals to

If $\sum_{k=1}^n \tan^{-1} \left( \frac{1}{k^2+k+1} \right) = \tan^{-1} ( \theta )$,then $\theta =$

$\sin ^{-1}\left(\frac{12}{13}\right)+\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{63}{16}\right)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module