Implicit Differentiation involving trigonometric functions.

Question

We are given the following condition:

$$\tan(x^3y^2)=6x^2+y^2$$

Find the derivative of $y$ w.r.t. $x$, i.e., find $y'=\dfrac{\textrm{d}y}{\textrm{d}x}$

I am having trouble getting started with this problem. Can someone please help me out?

Answer 1

Outlining steps with Chain Rule.

Differentiation of $ \tan \rightarrow \sec^2 = 1 + \tan^2 $ with respect to x:

$$\tan(x^3y^2)=6x^2+y^2$$

$$ [1 + (6 x^2+y^2)^2 ] ( 3 x^2 y^2 + x^3 2 y y^{'}) = ( 12 x + 2 y y^{'}) $$

Label quantity in square brackets as Q, collect $ y^{'} $ terms and simplify. HTH.

Answer 2

Using implicit differentiation, consider the function $$F=\tan(x^3y^2)-(6x^2+y^2)=0$$ and compute the partial derivatives. You get $$F'_x=3 x^2 y^2 \sec ^2\left(x^3 y^2\right)-12 x$$ $$F'_y=2 x^3 y \sec ^2\left(x^3 y^2\right)-2 y$$ and now $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}$$