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 $x^3 - 2x^2y^2 + 5x + y - 5 = 0$, then at $(1, 1)$, $y''_1 = $?

Find $\frac{dy}{dx}$ for the equation $2x + 3y = \sin y$.

If $y = \operatorname{Sin}^{-1}\left(\frac{4+5 \sin x}{5+4 \sin x}\right)$,then find $\frac{dy}{dx}$.

If $y=\sqrt{\cosh x+\sqrt{\cosh x+\dots}}$,then $\frac{d y}{d x}=$

Let $f(x), x \in [0, \infty)$ be a non-negative continuous function. If $f'(x) \cos x \le f(x) \sin x$ for all $x \ge 0$,then the value of $f(2\pi)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module