Let cos ^{-1}left(frac{y}{b}right)=log left(f | vedclass

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

Question

(A) Given: $\cos ^{-1}\left(\frac{y}{b}\right) = n \log \left(\frac{x}{n}\right)$.Taking the derivative with respect to $x$:$-\frac{1}{\sqrt{1 - (y/b)

Vedclass · Accepted Answer

$x^{2} y_{2}+x y_{1}+n^{2} y=0$

Vedclass · Answer

(A) Given: $\cos ^{-1}\left(\frac{y}{b}ight) = n \log \left(\frac{x}{n}ight)$.Taking the derivative with respect to $x$:$-\frac{1}{\sqrt{1 - (y/b)^2}} \cdot \frac{y_1}{b} = n \cdot \frac{n}{x} \cdot \frac{1}{n} = \frac{n}{x}$.$-\frac{1}{\sqrt{(b^2 - y^2)/b^2}} \cdot \frac{y_1}{b} = \frac{n}{x}$ $\Rightarrow -\frac{b}{\sqrt{b^2 - y^2}} \cdot \frac{y_1}{b} = \frac{n}{x}$.$-\frac{y_1}{\sqrt{b^2 - y^2}} = \frac{n}{x} \Rightarrow -x y_1 = n \sqrt{b^2 - y^2}$.Squaring both sides: $x^2 y_1^2 = n^2 (b^2 - y^2)$.Differentiating again with respect to $x$:$x^2 (2 y_1 y_2) + 2x y_1^2 = n^2 (-2 y y_1)$.Dividing by $2 y_1$ (assuming $y_1 
eq 0$):$x^2 y_2 + x y_1 = -n^2 y \Rightarrow x^2 y_2 + x y_1 + n^2 y = 0$.

Let $\cos ^{-1}\left(\frac{y}{b}\right)=\log \left(\frac{x}{n}\right)^{n}$. Then

Similar Questions

If $y=3 e^{2 x}+2 e^{3 x},$ prove that $\frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}+6 y=0$.

If $y=f(x)$ is a thrice differentiable function and a bijection,then $\frac{d^2 x}{d y^2}\left(\frac{d y}{d x}\right)^3+\frac{d^2 y}{d x^2}=$

If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$,then $\left[\left(1-x^2\right)^2 \frac{d^2 y}{d x^2}+y\left(1-x^2\right)\right] \frac{d y}{d x}=$

If $y = \sin(\sin x)$ and $y'' + f(x) \cdot y' + g(x) \cdot y = 0$,then $f(x) \cdot g(x) =$

Let $f(x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}$,$x \in [0, \pi] - \{\frac{\pi}{4}\}$. Then $f(\frac{7\pi}{12}) f''(\frac{7\pi}{12})$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module