If y = tan^{-1} sqrt{x^{2}-1},then the ratio | vedclass

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

Question

(B) Given $y = 	an^{-1} \sqrt{x^{2}-1}$. Let $x = \sec 	heta$,then $\sqrt{x^{2}-1} = 	an 	heta$. So,$y = 	an^{-1}(	an 	heta) = 	heta = \sec^{-

Vedclass · Accepted Answer

$\frac{1-2x^{2}}{x(x^{2}-1)}$

Vedclass · Answer

(B) Given $y = 	an^{-1} \sqrt{x^{2}-1}$. Let $x = \sec 	heta$,then $\sqrt{x^{2}-1} = 	an 	heta$. So,$y = 	an^{-1}(	an 	heta) = 	heta = \sec^{-1} x$. Now,differentiate with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(\sec^{-1} x) = \frac{1}{x \sqrt{x^{2}-1}}$. Now,find the second derivative $\frac{d^{2}y}{dx^{2}}$: $\frac{d^{2}y}{dx^{2}} = \frac{d}{dx} \left( (x(x^{2}-1)^{1/2})^{-1} ight) = -1 \cdot (x(x^{2}-1)^{1/2})^{-2} \cdot \frac{d}{dx} (x(x^{2}-1)^{1/2})$. Using the product rule: $\frac{d}{dx} (x(x^{2}-1)^{1/2}) = (x^{2}-1)^{1/2} + x \cdot \frac{1}{2}(x^{2}-1)^{-1/2} \cdot 2x = \sqrt{x^{2}-1} + \frac{x^{2}}{\sqrt{x^{2}-1}} = \frac{x^{2}-1+x^{2}}{\sqrt{x^{2}-1}} = \frac{2x^{2}-1}{\sqrt{x^{2}-1}}$. Thus,$\frac{d^{2}y}{dx^{2}} = - \frac{1}{x^{2}(x^{2}-1)} \cdot \frac{2x^{2}-1}{\sqrt{x^{2}-1}} = - \frac{2x^{2}-1}{x^{2}(x^{2}-1)^{3/2}}$. Finally,the ratio is: $\frac{d^{2}y/dx^{2}}{dy/dx} = \left( - \frac{2x^{2}-1}{x^{2}(x^{2}-1)^{3/2}} ight) \div \left( \frac{1}{x \sqrt{x^{2}-1}} ight) = - \frac{2x^{2}-1}{x^{2}(x^{2}-1)^{3/2}} \cdot x \sqrt{x^{2}-1} = - \frac{2x^{2}-1}{x(x^{2}-1)} = \frac{1-2x^{2}}{x(x^{2}-1)}$.

If $y = \tan^{-1} \sqrt{x^{2}-1}$,then the ratio $\frac{d^{2} y}{dx^{2}} : \frac{dy}{dx}$ is

Similar Questions

Let $f$ be a twice differentiable function such that $f^{\prime \prime}(x) = -f(x)$,$f^{\prime}(x) = g(x)$,and $h(x) = (f(x))^2 + (g(x))^2$. If $h(5) = 1$,then the value of $h(10)$ is

If $f$ is a polynomial,then the second derivative of $f(e^x)$ is

If $f(x) = e^{(x+1)^n}; (n \in N)$,then the value of $n$ for which $f''(1) = 67(2^n e^{2^n})$ is

The maximum value of $a$ such that the second derivative of $x^4+ax^3+\frac{3x^2}{2}+1$ is positive for all real $x$ is

If $y=5 \cos x-3 \sin x,$ prove that $\frac{d^{2} y}{d x^{2}}+y=0.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module