If y = fleft( frac{2x - 1}{x^2 + 1} right) an | vedclass

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

Question

(D) Given $y = f\left( \frac{2x - 1}{x^2 + 1} \right).$ Let $t = \frac{2x - 1}{x^2 + 1}.$By the chain rule,$\frac{dy}{dx} = f'(t) \cdot \frac{dt}{dx}.

Vedclass · Accepted Answer

$\frac{-2x^2 + 2x + 2}{(x^2 + 1)^2} \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2$

Vedclass · Answer

(D) Given $y = f\left( \frac{2x - 1}{x^2 + 1} \right).$ Let $t = \frac{2x - 1}{x^2 + 1}.$By the chain rule,$\frac{dy}{dx} = f'(t) \cdot \frac{dt}{dx}.$Since $f'(x) = \sin(x^2),$ we have $f'(t) = \sin(t^2) = \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2.$Now,calculate $\frac{dt}{dx}$ using the quotient rule:$\frac{dt}{dx} = \frac{(x^2 + 1)(2) - (2x - 1)(2x)}{(x^2 + 1)^2} = \frac{2x^2 + 2 - 4x^2 + 2x}{(x^2 + 1)^2} = \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2}.$Therefore,$\frac{dy}{dx} = \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2} \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2.$

If $y = f\left( \frac{2x - 1}{x^2 + 1} \right)$ and $f'(x) = \sin(x^2),$ then $\frac{dy}{dx} = $

Similar Questions

Find the derivative of the following function: $\frac{1+\frac{1}{x}}{1-\frac{1}{x}}$

Let $f(x) = x - \frac{1}{x}$,then $f^{\prime}(-1)$ is

If $\frac{d}{{dx}}\left[ {\frac{{2{x^3} + 3{x^2} + x - 3}}{{{x^2} + x - 2}}} \right] = A + \frac{B}{{{{(x - 1)}^2}}} + \frac{C}{{{{(x + 2)}^2}}}$ then $(A - B + C)$ is

If $f(2)=4$ and $f^{\prime}(2)=1$,then $\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$ is equal to

Find the derivative: $\frac{d}{dx} \left( \frac{1}{x^4 \sec x} \right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module