Finding the Second Derivative of a Function with Two Variables

[Maniac_XOX]

Homework Statement

find first and second partial derivative of z= tan (x^2*y^2)

Relevant Equations

tan (x^2*y^2)= sin(x^2*y^2)/cos(x^2*y^2)

Quotient rule: z= f/g ------ z'= (f'g - g'f)/g^2
starting with finding the derivative in respect to x, i treated y^2 as constant 'a': z'= [(a*2x*cos a*x^2)(sin a*x^2) - (- a*2x*sin ax^2)]/cos(a*x^2)^2=
[(a*2x*cos a*x^2)(sin a*x^2)+(a*2x*sin ax^2)]/cos(a*x^2)^2
For the derivative in respect to y it would be the same process, how do i go from here to finding the second derivative?

 

[stevendaryl]

Quotient rule: z= f/g ------ z'= (f'g - g'f)/g^2
starting with finding the derivative in respect to x, i treated y^2 as constant 'a': z'= [(a*2x*cos a*x^2)(sin a*x^2) - (- a*2x*sin ax^2)]/cos(a*x^2)^2=
[(a*2x*cos a*x^2)(sin a*x^2)+(a*2x*sin ax^2)]/cos(a*x^2)^2
For the derivative in respect to y it would be the same process, how do i go from here to finding the second derivative?

Well, you take the first derivative, and you have a new function of $x$ and $y$. What's the difficulty in taking a derivative of that? The second derivative uses the same principles as the first derivative.

 

[archaic]

In your case, I think you're not looking for the mixed partia derivatives, so go with the 1st and 4th formulas.
Source : https://www.khanacademy.org/math/mu...radient-articles/a/second-partial-derivatives

 

[LCKurtz]

@Maniac_XOX : Your post would be much more readable if you use the LaTeX typesetting that is available on this site:
https://www.physicsforums.com/help/latexhelp/

 

[Maniac_XOX]

@Maniac_XOX : Your post would be much more readable if you use the LaTeX typesetting that is available on this site:
https://www.physicsforums.com/help/latexhelp/

Thank you so much! didn't know we had such thing it's going to help a long way lol

 

[Maniac_XOX]

Well, you take the first derivative, and you have a new function of $x$ and $y$. What's the difficulty in taking a derivative of that? The second derivative uses the same principles as the first derivative.

You're right i was just unsure of the result I would get cos there is a lot of working out and wanted to be sure I made no arithmetic errors. I guess I just need to post the final answer and ask wether there are arithmetic errors. Thank you for your opinon x