If y=3x^2-2, then dy/d(2x-3)= ?
I've never taken derivatives with respect to anything other than a single variable, such as x, y or t, so i'm not sure how to start this problem! Thanks
how about let u = 2x-3
then 2x = u + 3
x = (u+3)/2
then y = 2(u+3)^2 /4 - 2
= (1/2)(u^2 + 6y + 9) - 2
dy/du = (1/2)(2u + 6) = u + 3
= (2x-3) + 3 = 2x
dy/d(2x-3) = 2x
or, you can let
u = 2x-3
dx/du = 1/2
dy/du = dy/dx * dx/du
= (6x)(1/2) = 3x
Reiny: there's a typo in your 2nd paragraph:
y = 3(u+3)^2-2
Thanks Steve, I see it now, used 2x^2 instead of 3x^2
To find the derivative of y with respect to (2x - 3), we can use the chain rule. The chain rule allows us to differentiate functions that are composed of one function within another.
Here are the steps to find dy/d(2x - 3):
1. Rewrite y = 3x^2 - 2 as a function of (2x - 3): y = 3(2x - 3)^2 - 2.
2. Let u = 2x - 3. Rewrite y in terms of u: y = 3u^2 - 2.
3. Differentiate y with respect to u: dy/du = 6u.
4. Use the chain rule to differentiate u with respect to x: du/dx = 2.
5. Multiply dy/du by du/dx to get dy/dx (which is what we want): dy/dx = (dy/du)(du/dx) = 6u * 2 = 12u.
6. Substitute the value of u back in: dy/dx = 12(2x - 3).
Therefore, the derivative of y = 3x^2 - 2 with respect to (2x - 3) is dy/d(2x - 3) = 12(2x - 3).