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Calculus with two variables.

Problem 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

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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.
 

LCKurtz

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@Maniac_XOX : Your post would be much more readable if you use the LaTeX typesetting that is available on this site:
 
@Maniac_XOX : Your post would be much more readable if you use the LaTeX typesetting that is available on this site:
Thank you so much! didn't know we had such thing it's gonna help a long way lol
 
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
 

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