Expert: Paul Klarreich - 12/18/2009

Question
To be honest, I'm not sure how to do the 2 problems I am about to send you at all. For the first one, I tried product rule, but my numbers got all mangled, and then I was confused if I had to take the square root of the final equation. For the second equation, I simply do not know how to do problems with 'e' in them.

I think I'd be able to learn from following each step, so this would really big a big help to me.

1. derivative of y^2=(2y+x^2)(x+y^2)

2. y = e^xsecx

Thank you for all your time, I appreciate it.
Carly

Answer
Questioner: carly
Country: United States
Category: Calculus
Private: No
Subject: derivatives
Question: To be honest, I'm not sure how to do the 2 problems I am about to send you at all. For the first one, I tried product rule, but my numbers got all mangled, and then I was confused if I had to take the square root of the final equation. For the second equation, I simply do not know how to do problems with 'e' in them.

I think I'd be able to learn from following each step, so this would really big a big help to me.

1. derivative of y^2=(2y+x^2)(x+y^2)

2. y = e^xsecx

Thank you for all your time, I appreciate it.
Carly
.......................................................

1. derivative of y^2=(2y+x^2)(x+y^2)

Yes, you need the product rule, but this is an exercise in implicit differentiation. (see previous anwsers.)

2y y' = (2y + x^2)(1 + 2y y') + (2y' + 2x)(x + y^2)

implify and solve for y'.  Yes, your answer will have x's and y's in it -- no way around it.

2 y y' = 2y + x^2 + 4y^2 y' + 2x^2 y y'  + 2x y' + 2y^2 y' + 2x^2 + 2xy^2

Bring all y' terms to one side, and the rest to the other side.  Solve for y'.

......................................

2. y = e^xsecx

If you have not learned how to handle  e^x, you will just have to wait until your class gets to it.