find dy/dx by implicit differentiation. x²y² + x sin(y) = 9 dy/dx = need help? read it watch it submit answer

Explanation:

Step1: Differentiate each term

Differentiate $x^{2}y^{2}$, $x\sin(y)$ and 9 with respect to $x$. For $x^{2}y^{2}$, use product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x^{2}$ and $v = y^{2}$. The derivative of $x^{2}$ with respect to $x$ is $2x$, and for $y^{2}$ using chain - rule we get $2y\frac{dy}{dx}$. So the derivative of $x^{2}y^{2}$ is $2xy^{2}+2x^{2}y\frac{dy}{dx}$. For $x\sin(y)$, using product - rule with $u = x$ and $v=\sin(y)$. The derivative of $x$ with respect to $x$ is 1, and the derivative of $\sin(y)$ with respect to $x$ is $\cos(y)\frac{dy}{dx}$. So the derivative of $x\sin(y)$ is $\sin(y)+x\cos(y)\frac{dy}{dx}$. The derivative of the constant 9 with respect to $x$ is 0. So, $\frac{d}{dx}(x^{2}y^{2}+x\sin(y))=\frac{d}{dx}(9)$ gives $2xy^{2}+2x^{2}y\frac{dy}{dx}+\sin(y)+x\cos(y)\frac{dy}{dx}=0$.

Step2: Isolate $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ on one side: $2x^{2}y\frac{dy}{dx}+x\cos(y)\frac{dy}{dx}=- 2xy^{2}-\sin(y)$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2x^{2}y + x\cos(y))=-2xy^{2}-\sin(y)$. Then $\frac{dy}{dx}=\frac{-2xy^{2}-\sin(y)}{2x^{2}y + x\cos(y)}$.

Answer:

$\frac{-2xy^{2}-\sin(y)}{2x^{2}y + x\cos(y)}$