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About Ahmed Salami
Expertise
I can provide good answers to questions dealing in almost all of mathematics especially from A`Level downwards. I believe i would be very helpful in calculus and can as well help a good deal in Physics with most emphasis directed towards mechanics.

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An engineering graduate. I have been doing maths and physics all my life.

 
   

You are here:  Experts > Teens > Homework/Study Tips > Calculus > Taking a derivative of a function in two variables

Calculus - Taking a derivative of a function in two variables


Expert: Ahmed Salami - 6/27/2009

Question
Hi. How do you take the derivative of a function in two variables?

For example how would I take the derivative dy/dx of the function x^2y+xy^3-3x=6 at x=4 ,y=1? Please let me know. Thank you.

Jason Goldman


Answer
Hi Jason,
The trick is to always add a dy/dx whenever u differentiate a function in y. I hope you know the product rule because we'll be applying it here.
Now,
(x^2)y + x(y^3) - 3x = 6
differentiating with respect to x,
[(x^2).1(dy/dx) + (2x)y] + [x.(3y^2)(dy/dx) + 1.(y^3)] - 3 = 0
(x^2)(dy/dx) + 2xy + (3xy^2)(dy/dx) + y^3 - 3 = 0
(x^2)(dy/dx) + (3xy^2)(dy/dx) = 3 - 2xy - y^3
[x^2 + 3xy^2](dy/dx) = 3 - 2xy - y^3
dy/dx = (3 - 2xy - y^3)/(x^2 + 3xy^2)
At x = 4, y = 1
dy/dx = (3 - 2.4.1 - 1^3)/(4^2 + 3.4.1^2)
     = -6/28
     = -3/14

This is the standard way. There is, however, another easier way using partial derivatives but i dont think you've started learning that.

Regards

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