When you have a trig function like tan^3[x(y^2)+y] and you need to take the derivative, do you still bring down the exponent on the tan function (3tan^2) before you continue?

Question

noelgreco · Answer

Yes.

It's actually easier to see if you use a form like this:

$$[\tan(xy ^{2}+y)]^{3}$$

anonymous · Answer

Afterward would you leave it as $$3[\tan(xy^2+y)]^2$$ and keep going?

noelgreco · Answer

Yep - chain rule, implicit differentiation, and you're there.

anonymous · Answer

Now when doing the implicit differentiation, do you write: $$d/dx[3\tan(xy^2+y)^2]=dx/dx (x)$$ or $$dy/dx[3\tan(xy^2+y)^2]=dx/dx(x)$$

noelgreco · Answer

$$3[\tan(xy ^{2}+y]^{2}\frac{ dy }{ dx }\tan(xy ^{2}+y)$$

noelgreco · Answer

That should be d/dx not dy/dx

anonymous · Answer

Ok cool, it kept throwing me off. Thanks!

anonymous · Answer

Ok Im working through it and as Im differentiating, when I factor out the dy/dx, that d/dx gets pulled out also?

noelgreco · Answer

No, that just indicates you must take the derivative of the statement after.  The next step would be$$\sec ^{2}(xy ^{2}+y)$$
times the derivative of
$$(xy ^{2} +y)$$