implicit

[mikeb]

find dy/dx by implicit differentiation:

tan ^3 (xy^2 +y)= x

i think im missing a step, can you help?

 

[tkhunny]

find dy/dx by implicit differentiation:
tan^3 (xy^2 +y)= x
i think im missing a step, can you help?

It would be helpful to provide the steps you are not missing.

Calculus Step - Serious Chain Rule Application
Three Pieces
1) tan^3 --> 3*tan^2
2) tan --> sec^2
3) argument

3*tan^2 (xy^2 +y)*sec^2(xy^2 +y)*(x*2y*y' + y^2 + y') = 1

The rest is algebra.

 

[mikeb]

where did the f' come from?

 

[mikeb]

im sorry, i mean y'

 

[tkhunny]

It is the crux of implicit differentiation. We are assuming that 'y' is SOME function of 'x'. We just don't know what function. When it comes to finding it's derivative, we don't know how to do it, so we just write y' and worry about it later.

(d/dx)(x*y) = x*y' + y*x' = x*y' + y