Question

Need help with implicit differentiation

Answer 1

\[\tan^{-1} (x^2 y)=x+xy^2\]

Answer 2

derivative of inverse tangent is 1/1+x^2 right

Answer 3

Well if you're doing implicit differentiation I think you should make it easy on yourself and do implicit differentiation on this:

\[x^2y = \tan (x+xy^2)\]

You can, perhaps ironically, derive the derivative of inverse tangent with implicit differentiation.

Answer 4

I say that cause I don't remember if that's true or not, so I guess I'll just show it and figure it out real quick:

\[y = \tan^{-1} x\]
\[\tan y = x\]
\[\sec^2 y \frac{dy}{dx} = 1\]
\[\frac{dy}{dx} = \frac{1}{\sec^2 y} = \cos^2 y\]
We know since \the second line there is \(\tan y = x\) that we have:
We can solve for the hypotenuse with the pythagorean theorem, \(\sqrt{1+x^2}\)
Which we need cause \(\cos y\) is adjacent over hypotenuse, so:

\[\frac{dy}{dx} = \left( \frac{1}{\sqrt{1+x^2}} \right)^2 = \frac{1}{1+x^2}\]
Which is indeed what you guessed, so good memory!

Answer 5

Don't be afraid to ask questions, I sorta just rushed through this and probably left out some critical steps cause I literally forget this all the time so I'm used to doing this exact derivation haha.

Answer 6

so plug in x^2y in place of x?

Answer 7

yes