Find the equation of the tangent line tangent to the given point on the following curve.

Question

It is clear that this graph passes through (2,1). But is it possible to find the equation of the tangent line at this point by differentiating it? If so, how can it be differentiated?

Answer 1

Taking the derivative of the equation with respect to $x$, we have $$e^{x^3-2x^2y^2}\left(3x^2-\left(4xy^2+4x^2y\frac{dy}{dx}\right)\right)-\sec^2(2y^3-xy^2)\left(6y^2\frac{dy}{dx}-\left(y^2+2xy\frac{dy}{dx}\right)\right)=0.$$

From here, you can plug in the point $(2,1)$ into the equation and solve for $\frac{dy}{dx}$. Once you have $\frac{dy}{dx}$, this is the slope of your line, so you can use point-slope form to write the equation for your line.

This method of differentiating is known as implicit differentiation. The key is that whenever you differentiate an expression in $y$ (with respect to $x$), you must multiply by $\frac{dy}{dx}$ due to the chain rule.