Is this chain rule correct?

Question

Let f(x, y) be a differentiable function from R^2 -> R, and let g(t) = (x(t), y(t)) be a differentiable function from R -> R^2.

Then the derivative of f(g(t)) is (f_x dx/dt, f_y dy/dt).

(I'm feeling low about my multivariable calculus knowledge.)

I'm asking if my statement "Then the derivative of..." is correct.

Prince Ali: f(x) = (x, 5) means a function from R -> R^2, where the value of the first coordinate is x, and the value of the second coordinate is 5.

Answer 1

The comma doesn't make sense. What does a comma mean in a function? If I told you f(x) = x,5 what does that mean.

Doug's answer is correct, you take the derivates with respect to each variable multiply by it d whatever it is and then add them all.

Answer 2

If you have

z = f(g(t) = f(x(t),y(t))

then to get dz/dt you need to do partial derivatives. The partial derivative operator (the one that looks like a 'backwards' 6) isn't in the ASCII character set, so I'll use the lowercase delta (δ) to represent it. I hope it doesn't confuse you (or me )

If you have z = f(g(t)) then

dz/dt = f'(g(t))∙g'(t)

just like you'd expect. To get g'(t) let w = g(x(t),y(t)) and form its derivative as

dw/dt = (δw/δx)(dx/dt) + (δw/δy)(dy/dt) = g'(t)

Note that the dx/dt and dy/dt actually mean (d/dt)x(t) and (d/dt)y(t)

so the final equation is

dz/dt = f'(g(t))∙((δw/δx)(dx/dt) + (δw/δy)(dy/dt))

Kinda messy, but there it is


Doug

For 'Prince Ali'

The comma serves to seperate the variables in a function of more than one variable. For example, if you define

f(x,y) = x² - y²

then ask for f(x(t),y(t) with x(t) = t² + 1 and y(t) = t² - 1 then you'd have

f(x(t),y(t) = (t² + 1)² - (t² - 1)²

d/dt is an 'operator notation'. It means "take the derivative (of what follows) with respect to t."

The partial derivative δw/δx (where w is a function of two or more variables) means "take the derivative of w with respect to x treating all of the other variables as if they were constants." For example, if

w(x,y) = x² + xy then

δw/δx = 2x + y

and

δw/δy = x

Essentially you're taking the derivative of the function along one axis by holding the other variable(s) constant.


Doug

Answer 3

The chain rule for single variable calculus is:

f'(g(x))=f'(g(x))g'(x).

I don't know about multivariable, though.

Answer 4

Doug's answer is correct.

Answer 5

can you give a specific instance,a sum maybe illustrating what you are trying to say?