Question

tan^-1(x^2y)=x+xy^2

Answer 1

find dy/dx by implicit differentation

Answer 2

i don't get what to do with the left part

Answer 3

First, a bit of review:
What is the derivative of the inverse tangent function?

Answer 4

sec2x

Answer 5

right?

Answer 6

x^(2y) is, I assume, the argument of the inverse tangent function. Is that correct?

to answer your question: 1) sec2x means sec (2x). No.
What you need and want is 2) (sec x)^2. Big difference.

Answer 7

its x^2*y

Answer 8

Your left side, tan^-1(x^2y), apparently stems from:\[\tan ^{-1}x ^{2y}\]
Am I correct here? If not, DRAW the problem statement.

Answer 9

yes thats correct

Answer 10

Please note that what I typed is not the same as your

x^2y. It's worth spending the time to check and double check that you have presented the original problem correctly. I'd suggest you learn how to use equation Editor and Draw, if you haven't already.

Answer 11

so now we've reviewed the derivative of tan x, and the correct result is (sec x)^2.

Answer 12

That's the derivative with respect to x.

Now suppose instead that you want the derivative with respect to x of tan u. What would that derivative look like?

Answer 13

Hint: because 'u' is a separate function of x, we MUST use the Chain Rule here.

Answer 14

just looked through my notes and found that the derivative of tan^-1x is 1/1+x^2

Answer 15

I'm very sorry; you are right.

Answer 16

so i multiply that times x^(2)*y

Answer 17

\[\frac{ d }{ dx }\tan ^{-1}x=\frac{ 1 }{ 1+x^2 }\]