Question

Can anyone help me implicitly differentiate this??

X^3+xtan^-1(y)=e^y

Answer 1

What is giving you a problem. Use normal differentiation rules and remember chain rule:
[g(y)]'=g'(y)y'

Answer 2

\[x^3+x \tan^{-1} (y)=e^y\]

Differentiating I got:

3x^2+x/(1+y^2)(dy/dx)+tan^(-1)(y)=e^y(dy/dx)

Answer 3

Then to simplify:

3x^2+tan^(-1)(y)=e^y(dy/dx)-x/(1+y^2)(dy/dx)

Factoring out dy/dx of RHS

3x^2+tan^(-1)(y)=dy/dx(e^y-(x/(1+y^2))

Solving for dy/dx

3x^2+tan^(-1)(y)/(e^y-x/(1+y^2)=dy/dx

Answer 4

sorry meant the first one from left to right

Answer 5

That is what I am having trouble with - why should it be in the numerator?

dy/dx of arctan(x) = 1/(x^2+1)

Right?

Answer 6

Let g(y)=arctan(y)
We know [g(y)]'=g(y)*(dy/dx) by chain rule. So we have \[\frac{dy}{dx}\frac{1}{1+y^{2}}\]
The dy/dx is technically in the numerator of the fraction

Answer 7

Sorry [g(y)]'=g'(y)*(dy/dx)

Answer 8

So this is what it should look like?

(dy/dx)/1/(1+y^2)

(dy/dx)*(1+y^2)/1

(1+y^2)dy/dx

(not forgetting to multiply by x in my final answer because of the product rule)

Answer 9

No no you don't need to manipulate it. The derivative of arctan(y) is just as show above