If x^2+y^2+sin y=4,then the value of frac{d^2 | vedclass

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

Question

(B) Given the equation $x^2+y^2+\sin y=4$. At $x=-2$,we have $(-2)^2+y^2+\sin y=4$,which simplifies to $4+y^2+\sin y=4$,so $y^2+\sin y=0$. Since $y=0$

Vedclass · Accepted Answer

$-34$

Vedclass · Answer

(B) Given the equation $x^2+y^2+\sin y=4$. At $x=-2$,we have $(-2)^2+y^2+\sin y=4$,which simplifies to $4+y^2+\sin y=4$,so $y^2+\sin y=0$. Since $y=0$ is a solution,we have $y=0$ at $x=-2$. Differentiating $x^2+y^2+\sin y=4$ with respect to $x$,we get $2x+2y\frac{dy}{dx}+\cos y \frac{dy}{dx}=0$. Substituting $x=-2$ and $y=0$,we get $2(-2)+2(0)\frac{dy}{dx}+\cos(0)\frac{dy}{dx}=0$,which gives $-4+\frac{dy}{dx}=0$,so $\frac{dy}{dx}=4$. Differentiating $2x+(2y+\cos y)\frac{dy}{dx}=0$ again with respect to $x$,we get $2+(2\frac{dy}{dx}-\sin y \frac{dy}{dx})\frac{dy}{dx}+(2y+\cos y)\frac{d^2y}{dx^2}=0$. Substituting $x=-2, y=0, \frac{dy}{dx}=4$,we get $2+(2(4)-\sin(0)(4))(4)+(2(0)+\cos(0))\frac{d^2y}{dx^2}=0$. This simplifies to $2+(8)(4)+(1)\frac{d^2y}{dx^2}=0$,so $2+32+\frac{d^2y}{dx^2}=0$. Thus,$\frac{d^2y}{dx^2}=-34$.

If $x^2+y^2+\sin y=4$,then the value of $\frac{d^2 y}{d x^2}$ at $x=-2$ is

Similar Questions

If $f(x) = (x^2 - 1)^7$, then $f^{(14)}(x)$ is equal to

If $y=a \cos 3 x+b e^{-x}$, then $y^{\prime \prime}(3 \sin 3 x-\cos 3 x)=$

If $y_k$ is the $k$-th derivative of $y$ with respect to $x$,and $y = \cos(\sin x)$,then $y_1 \sin x + y_2 \cos x$ is equal to

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module