frac{d}{d x}left(cos ^{-1}left(frac{4 x^3}{27 | vedclass

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

Question

(C) Let $y = \cos ^{-1}\left(\frac{4 x^3}{27}-x\right)$.We can rewrite the expression as $y = \cos ^{-1}\left(4\left(\frac{x}{3}\right)^3 - 3\left(\fr

Vedclass · Accepted Answer

$\frac{-3}{\sqrt{9-x^2}}$

Vedclass · Answer

(C) Let $y = \cos ^{-1}\left(\frac{4 x^3}{27}-xight)$.We can rewrite the expression as $y = \cos ^{-1}\left(4\left(\frac{x}{3}ight)^3 - 3\left(\frac{x}{3}ight)ight)$.Let $\frac{x}{3} = \cos A$,then $A = \cos ^{-1}\left(\frac{x}{3}ight)$.Substituting this into the expression,we get $y = \cos ^{-1}(4 \cos ^3 A - 3 \cos A)$.Using the trigonometric identity $\cos(3A) = 4 \cos ^3 A - 3 \cos A$,we have $y = \cos ^{-1}(\cos 3A) = 3A$.Thus,$y = 3 \cos ^{-1}\left(\frac{x}{3}ight)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 3 	imes \left(-\frac{1}{\sqrt{1 - (x/3)^2}}ight) 	imes \frac{d}{dx}\left(\frac{x}{3}ight)$.$\frac{dy}{dx} = 3 	imes \left(-\frac{1}{\sqrt{(9-x^2)/9}}ight) 	imes \frac{1}{3} = 3 	imes \left(-\frac{3}{\sqrt{9-x^2}}ight) 	imes \frac{1}{3} = -\frac{3}{\sqrt{9-x^2}}$.

$\frac{d}{d x}\left(\cos ^{-1}\left(\frac{4 x^3}{27}-x\right)\right)=$

Similar Questions

$\frac{d}{dx} \left( \tan^{-1} \frac{x}{\sqrt{a^2 - x^2}} \right) = $

If $y(x)=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$ and $\left(1+a^2 x^2\right) y^{\prime \prime}+g(x) y^{\prime}=0$, then the sum of the roots of the equation $1+a^2 x^2+g(x)=0$ is

If $f(x) = \cos^{-1} \left[ \frac{1 - (\log x)^2}{1 + (\log x)^2} \right]$,then $f'(e) = \_\_\_\_$

Differentiate the following with respect to $x$: $\sin ^{-1}\left(\frac{2^{x+1}}{1+4^{x}}\right)$

If $y=\sin ^{-1}\left[x \sqrt{1-x^2}-\sqrt{x} \sqrt{1-x}\right]$ and $0 < x < 1$,then $\frac{d y}{d x}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module