y= u^5 +u^3

u= 3/v - 4v

v= 3 - x^2

Use the chain rule in Leibniz notation to find (dy/dx) when x = 2

Text gives the solution: -48608

do i use:

(dy/dx) = (dy/du)(du/dv)(dv/dx) because when I do, I don't get the right answer.

Printable View

- May 1st 2008, 07:25 PMwhite_capEasy question: i'm new too calc. (urgent)
y= u^5 +u^3

u= 3/v - 4v

v= 3 - x^2

Use the chain rule in Leibniz notation to find (dy/dx) when x = 2

Text gives the solution: -48608

do i use:

(dy/dx) = (dy/du)(du/dv)(dv/dx) because when I do, I don't get the right answer. - May 1st 2008, 07:50 PMMr. Engineer
- May 1st 2008, 07:52 PMo_O
Well take the derivative:

(Just the application of the power rule over and over again)

Now use x = 2 to find v(2) and u(2):

Then plug all these values into your expression for at x = 2