Question

find $\frac{dy}{dx}$ by implicit differentiation.
$7ysin(x^{2}) = 9xsin(y^{2})$
$\frac{dy}{dx}=$

Explanation:

Step1: Differentiate both sides

Differentiate $7y\sin(x^{2})$ and $9x\sin(y^{2})$ with respect to $x$ using product - rule $(uv)' = u'v+uv'$ and chain - rule.
The derivative of $7y\sin(x^{2})$ is $7\frac{dy}{dx}\sin(x^{2})+7y\cdot2x\cos(x^{2})$.
The derivative of $9x\sin(y^{2})$ is $9\sin(y^{2})+9x\cdot2y\cos(y^{2})\frac{dy}{dx}$.
So we have $7\frac{dy}{dx}\sin(x^{2}) + 14xy\cos(x^{2})=9\sin(y^{2})+18xy\cos(y^{2})\frac{dy}{dx}$.

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

Move all terms with $\frac{dy}{dx}$ to one side:
$7\frac{dy}{dx}\sin(x^{2})-18xy\cos(y^{2})\frac{dy}{dx}=9\sin(y^{2}) - 14xy\cos(x^{2})$.
Factor out $\frac{dy}{dx}$:
$\frac{dy}{dx}(7\sin(x^{2})-18xy\cos(y^{2}))=9\sin(y^{2}) - 14xy\cos(x^{2})$.
Then $\frac{dy}{dx}=\frac{9\sin(y^{2}) - 14xy\cos(x^{2})}{7\sin(x^{2})-18xy\cos(y^{2})}$.

Answer:

$\frac{9\sin(y^{2}) - 14xy\cos(x^{2})}{7\sin(x^{2})-18xy\cos(y^{2})}$