If (a+sqrt{2} b cos x)(a-sqrt{2} b cos y)=a^2 | vedclass

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

Question

(B) Given the equation: $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2$. Differentiating both sides with respect to $x$: $(a+\sqrt{2} b \cos x) \

Vedclass · Accepted Answer

$\frac{a-b}{a+b}$

Vedclass · Answer

(B) Given the equation: $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2$. Differentiating both sides with respect to $x$: $(a+\sqrt{2} b \cos x) \cdot (\sqrt{2} b \sin y \frac{dy}{dx}) + (a-\sqrt{2} b \cos y) \cdot (-\sqrt{2} b \sin x) = 0$. At the point $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$,we have $\cos x = \cos y = \frac{1}{\sqrt{2}}$ and $\sin x = \sin y = \frac{1}{\sqrt{2}}$. Substituting these values: $(a+\sqrt{2} b \cdot \frac{1}{\sqrt{2}}) \cdot (\sqrt{2} b \cdot \frac{1}{\sqrt{2}} \cdot \frac{dy}{dx}) + (a-\sqrt{2} b \cdot \frac{1}{\sqrt{2}}) \cdot (-\sqrt{2} b \cdot \frac{1}{\sqrt{2}}) = 0$. $(a+b) \cdot (b \frac{dy}{dx}) + (a-b) \cdot (-b) = 0$. $(a+b) b \frac{dy}{dx} = b(a-b)$. Since $b > 0$,we can divide by $b$: $(a+b) \frac{dy}{dx} = a-b$. Therefore,$\frac{dy}{dx} = \frac{a-b}{a+b}$.

If $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2$ where $a>b>0$,then at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$,$\frac{dy}{dx}=$

Similar Questions

If $\sqrt{1-x^6}+\sqrt{1-y^6}=a(x^3-y^3)$,then $y^2 \frac{dy}{dx}=$

If $x^2 \tan ^{-1} \frac{y}{x}-y^2 \tan ^{-1} \frac{x}{y}=k$,then $\left(\frac{d y}{d x}\right)_{(1,1)}=$

If $f(x)$ is an invertible and twice differentiable function satisfying $f'(x) = \int_{0}^{f(x)} f^{-1}(t) dt$ for all $x \in R$ and $f'(0) = 1$,then $f'(1)$ is equal to:

If the slope of a tangent to the curve $xy + ax + by = 0$ at the point $(1, 1)$ on it is $2$,then $a - b$ is

If $2^x+2^y=2^{x+y}$, then $\frac{d y}{d x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module