Find the derivative: frac{d}{dx} tan^{-1} lef | vedclass

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

Question

(D) Let $y = 	an^{-1} \left( \frac{ax - b}{bx + a} ight)$.We can rewrite the argument of the inverse tangent function by dividing the numerator and

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let $y = 	an^{-1} \left( \frac{ax - b}{bx + a} ight)$.We can rewrite the argument of the inverse tangent function by dividing the numerator and denominator by $a$:$y = 	an^{-1} \left( \frac{x - \frac{b}{a}}{1 + \frac{b}{a}x} ight)$.Using the identity $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1} \left( \frac{A - B}{1 + AB} ight)$,we get:$y = 	an^{-1}(x) - 	an^{-1} \left( \frac{b}{a} ight)$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left( 	an^{-1}(x) ight) - \frac{d}{dx} \left( 	an^{-1} \left( \frac{b}{a} ight) ight)$.Since $	an^{-1} \left( \frac{b}{a} ight)$ is a constant,its derivative is $0$.Therefore,$\frac{dy}{dx} = \frac{1}{1 + x^2} - 0 = \frac{1}{1 + x^2}$.Comparing this with the given options,the correct answer is not listed.

Find the derivative: $\frac{d}{dx} \tan^{-1} \left( \frac{ax - b}{bx + a} \right) = $

Similar Questions

Let $f(x) = \mathop {\text{Lim}}\limits_{h \to 0} \frac{1}{h} \int\limits_x^{x + h} \frac{dt}{t + \sqrt{1 + t^2}}$,then $\mathop {\text{Lim}}\limits_{x \to -\infty} x \cdot f(x)$ is

If $f(1)=1$ and $f^{\prime}(1)=3$,then the derivative of $f(f(f(x)))+(f(x))^2$ at $x=1$ is

If $f(x) = 4x^3 + 3x^2 + 3x + 4$,$x \neq 0$,then $\frac{d}{dx}\left(x^3 \cdot f\left(\frac{1}{x}\right)\right) =$ . . . . . .

If $y = \sin [\cos (\sin x)],$ then $dy/dx = $

If $f$ and $g$ are differentiable functions such that $g'(a) = 2$ and $g(a) = b$,and if $f \circ g$ is an identity function,then $f'(b)$ has the value equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module