tan^3 (x y^2+y)=x use implicit differentiation

Question

 Find Dy/dx?

Answer 1

Juts take derivative of tan³(xy²+y)=x and keep in mind that y is a function of x: y=y(x).

(tan³(xy²+y)'=x'

(Use Chain rule several times)

3tan²(xy²+y)(tan(xy²+y))' =1

3tan²(xy²+y)1/(cos²(xy²+y))(xy²+y)' =1

3tan²(xy²+y)1/(cos²(xy²+y))(y²+2xyy'+y') =1

3tan²(xy²+y)1/(cos²(xy²+y))(y²+(2xy+1)y') =1

If you'd like, you can express y' as a function f(x,y).

In general the implicit derivative for the function F(x, y) can be expressed as  y'=-F_x(x,y)/F_y(x,y). You can try this way as well for

F(x, y)= tan³(xy²+y)-x

Answer 2

the first stage of differentiating w.r.t.x gives this:

3tan²(xy²+y)sec²(xy²+y)[y²+2xyy'+y'] = 1   where, of course, y' represents dy/dx.

 

Now you distribute and then gather together all terms having y' as a factor, and segregate other terms to opposite side, according to rules of Algebra. Your coefficient of y' will be nasty, and something that I surely do not wish to type, but it's easy on paper from this point onward.

 

Be sure to check all of the work.