AboutAhmed 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.
Experience An engineering graduate. I have been doing maths and physics all my life.
Question 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
---------- FOLLOW-UP ----------
QUESTION: Hi and thank you. I haven't done anything with Calculus for years, I know this-I just needed a refreshing. What's the way with partial derivatives? Please tell me.
Jason Goldman
Answer Hi Jason,
By the method of partial derivatives,
dy/dx = -(&z/dx)/(&z/dy)
where & represents the partial derivative operator as different from the ordinary operator d.
&z/dx and &z/dy are the partial derivatives with respect to x and y respectively. The former is taking y to be a constant and the second is vice versa.
The trick is to take z to be 0.
For (x^2)y + x(y^3) - 3x = 6
(x^2)y + x(y^3) - 3x - 6 = 0
and so,
z = (x^2)y + x(y^3) - 3x - 6
&z/dx = 2xy + y^3 - 3
&z/dy = x^2 + 3xy^2
Therefore,
dy/dx = -(2xy + y^3 - 3)/(x^2 + 3xy^2)
= (3 - 2xy - y^3)/(x^2 + 3xy^2)
as before.
