Question

Differentiate
tan((x^2) y-x) = x

Answer 1

$\frac{d(-tan(x-x^{2}y(x)))}{dx}=\frac{d(x)}{dx}$

Factor out constants $-\frac{d(tan(x-x^{2}y(x)))}{dx}=\frac{d(x)}{dx}$

$\frac{d(x)}{dx}=1$

Using the chain rule,

$\frac{d(tan(x-x^{2}y(x)))}{dx}=\frac{du}{dx}*sec^{2}(u)$ where $u=x-x^{2}y(x)$

and

$\frac{dtan(u)}{du}=sec^{2}(u)$

we get

$(-1)*(sec^{2}(x-x^{2}y(x))\frac{d(x-x^{2}y(x))}{dx}=1$

$(-1)sec^{2}(x-x^{2}y(x))(\frac{d(x)}{dx}-\frac{d(x^{2}y(x))}{dx})=1$

$-sec^{2}(x-x^{2}y(x))(1-\frac{d(x^{2}y(x))}{dx})=1$

Use the product rule

$\frac{d(uv)}{dx}=v*\frac{du}{dx}+u*\frac{dv}{dx}$

where $u=x^{2}$, $v=y(x)$

$-sec^{2}(x-x^{2}y(x))(-y(x)*\frac{d(x^{2})}{dx}-x^{2}y^{\prime}(x)+1)=1$

Then we get

$-(-x^{2}y^{\prime}(x)-2xy(x)+1)sec^{2}(x-x^{2}y(x))=1$

and finaly

$y^{\prime}(x)=(cos^{2}(x-x^{2}y)-2xy+1)/x^{2}$