Question

find $\frac{dy}{dx}$ by implicit differentiation. $\tan^{-1}(2x^{2}y)=x + 4xy^{2}$ $\frac{dy}{dx}=$ blank resources read it watch it 8. -/1 points find $\frac{dy}{dx}$ by implicit differentiation. $xsin(y)+ysin(x)=2$ $\frac{dy}{dx}=$ blank resources read it

Explanation:

Step1: Differentiate left - hand side

Differentiate $\tan^{-1}(2x^{2}y)$ with respect to $x$ using the chain rule. The derivative of $\tan^{-1}(u)$ is $\frac{1}{1 + u^{2}}\cdot\frac{du}{dx}$, where $u = 2x^{2}y$. So, $\frac{d}{dx}\tan^{-1}(2x^{2}y)=\frac{1}{1+(2x^{2}y)^{2}}\cdot(4xy + 2x^{2}\frac{dy}{dx})=\frac{4xy + 2x^{2}\frac{dy}{dx}}{1 + 4x^{4}y^{2}}$.

Step2: Differentiate right - hand side

Differentiate $x+4xy^{2}$ with respect to $x$. The derivative of $x$ is $1$, and for $4xy^{2}$ use the product rule $(uv)^\prime=u^\prime v+uv^\prime$ where $u = 4x$ and $v = y^{2}$. So, $\frac{d}{dx}(x + 4xy^{2})=1+4y^{2}+8xy\frac{dy}{dx}$.

Step3: Solve for $\frac{dy}{dx}$

Set the two derivatives equal: $\frac{4xy + 2x^{2}\frac{dy}{dx}}{1 + 4x^{4}y^{2}}=1 + 4y^{2}+8xy\frac{dy}{dx}$. Cross - multiply: $4xy + 2x^{2}\frac{dy}{dx}=(1 + 4y^{2}+8xy\frac{dy}{dx})(1 + 4x^{4}y^{2})$. Expand the right - hand side: $4xy + 2x^{2}\frac{dy}{dx}=1 + 4x^{4}y^{2}+4y^{2}+16x^{4}y^{4}+8xy\frac{dy}{dx}+32x^{5}y^{3}\frac{dy}{dx}$. Rearrange terms to isolate $\frac{dy}{dx}$: $2x^{2}\frac{dy}{dx}-8xy\frac{dy}{dx}-32x^{5}y^{3}\frac{dy}{dx}=1 + 4x^{4}y^{2}+4y^{2}+16x^{4}y^{4}-4xy$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2x^{2}-8xy - 32x^{5}y^{3})=1 + 4x^{4}y^{2}+4y^{2}+16x^{4}y^{4}-4xy$. Then $\frac{dy}{dx}=\frac{1 + 4x^{4}y^{2}+4y^{2}+16x^{4}y^{4}-4xy}{2x^{2}-8xy - 32x^{5}y^{3}}$.

for second part:

Step1: Differentiate left - hand side

Differentiate $x\sin(y)+y\sin(x)$ with respect to $x$ using the product rule. For $x\sin(y)$, it is $\sin(y)+x\cos(y)\frac{dy}{dx}$, and for $y\sin(x)$ it is $\frac{dy}{dx}\sin(x)+y\cos(x)$. So, $\frac{d}{dx}(x\sin(y)+y\sin(x))=\sin(y)+x\cos(y)\frac{dy}{dx}+\frac{dy}{dx}\sin(x)+y\cos(x)$.

Step2: Differentiate right - hand side

The derivative of the constant $2$ is $0$.

Step3: Solve for $\frac{dy}{dx}$

Set $\sin(y)+x\cos(y)\frac{dy}{dx}+\frac{dy}{dx}\sin(x)+y\cos(x)=0$. Rearrange terms to isolate $\frac{dy}{dx}$: $x\cos(y)\frac{dy}{dx}+\frac{dy}{dx}\sin(x)=-y\cos(x)-\sin(y)$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(x\cos(y)+\sin(x))=-y\cos(x)-\sin(y)$. Then $\frac{dy}{dx}=-\frac{y\cos(x)+\sin(y)}{x\cos(y)+\sin(x)}$.

Answer:

$\frac{1 + 4x^{4}y^{2}+4y^{2}+16x^{4}y^{4}-4xy}{2x^{2}-8xy - 32x^{5}y^{3}}$