If y=cos ^{-1}left(frac{6 x-2 x^2-4}{2 x^2-6 | vedclass

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

Question

(D) Given $y=\cos ^{-1}\left(\frac{6 x-2 x^2-4}{2 x^2-6 x+5}\right)$.Let $v = \frac{6 x-2 x^2-4}{2 x^2-6 x+5}$.Note that $v = \frac{-(2x^2-6x+5)+1}{2x

Vedclass · Accepted Answer

$\frac{2}{2 x^2-6 x+5}$

Vedclass · Answer

(D) Given $y=\cos ^{-1}\left(\frac{6 x-2 x^2-4}{2 x^2-6 x+5}\right)$.Let $v = \frac{6 x-2 x^2-4}{2 x^2-6 x+5}$.Note that $v = \frac{-(2x^2-6x+5)+1}{2x^2-6x+5} = -1 + \frac{1}{2x^2-6x+5}$.Using the chain rule, $\frac{dy}{dx} = \frac{-1}{\sqrt{1-v^2}} \cdot \frac{dv}{dx}$.Calculating $\frac{dv}{dx}$ using the quotient rule:$\frac{dv}{dx} = \frac{(6-4x)(2x^2-6x+5) - (6x-2x^2-4)(4x-6)}{(2x^2-6x+5)^2} = \frac{2(3-2x)(2x^2-6x+5) + 2(2x^2-6x+4)(2x-3)}{(2x^2-6x+5)^2} = \frac{2(2x-3)[-(2x^2-6x+5) + (2x^2-6x+4)]}{(2x^2-6x+5)^2} = \frac{2(2x-3)(-1)}{(2x^2-6x+5)^2} = \frac{-2(2x-3)}{(2x^2-6x+5)^2}$.Now, $1-v^2 = 1 - \left(\frac{6x-2x^2-4}{2x^2-6x+5}\right)^2 = \frac{(2x^2-6x+5)^2 - (6x-2x^2-4)^2}{(2x^2-6x+5)^2} = \frac{(2x^2-6x+5 - 6x+2x^2+4)(2x^2-6x+5 + 6x-2x^2-4)}{(2x^2-6x+5)^2} = \frac{(4x^2-12x+9)(1)}{(2x^2-6x+5)^2} = \frac{(2x-3)^2}{(2x^2-6x+5)^2}$.Thus, $\sqrt{1-v^2} = \frac{|2x-3|}{2x^2-6x+5}$.Substituting back: $\frac{dy}{dx} = \frac{-1}{\frac{|2x-3|}{2x^2-6x+5}} \cdot \frac{-2(2x-3)}{(2x^2-6x+5)^2} = \frac{2(2x-3)}{|2x-3|(2x^2-6x+5)}$.Assuming $2x-3 > 0$, $\frac{dy}{dx} = \frac{2}{2x^2-6x+5}$.

If $y=\cos ^{-1}\left(\frac{6 x-2 x^2-4}{2 x^2-6 x+5}\right)$, then $\frac{d y}{d x}=$

Similar Questions

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

The differential coefficient of ${\tan ^{ - 1}}\left( {\frac{x}{{1 + \sqrt {1 - {x^2}} }}} \right)$ with respect to ${\sin ^{ - 1}}x$ is:

If $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1+9^x}\right)$,then $f^{\prime}\left(\frac{1}{2}\right)$ equals

Let $f : R \rightarrow R$ be a differentiable function and $f(1) = 4$. Then the value of $\lim_{x \rightarrow 1} \int_{4}^{f(x)} \frac{2t \, dt}{x - 1}$ is:

If $y=\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)$,then $\frac{d y}{d x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module