# Math Help - Applying chain rule

1. ## Applying chain rule

Find the derivative of the following:

$y = (2x - 5)(4 - x)^{-1}$

I got $y' = 2x(4 - x)^{-1} + (2x - 5)(4 - x)^{-2}$ or $y' = \frac{2x}{4 - x} + \frac{2x - 5}{(4 - x)^2}$

Correct?

2. Originally Posted by VitaX
Find the derivative of the following:

$y = (2x - 5)(4 - x)^{-1}$

I got $y' = 2x(4 - x)^{-1} + (2x - 5)(4 - x)^{-2}$ or $y' = \frac{2x}{4 - x} + \frac{2x - 5}{(4 - x)^2}$

Correct?
$y' = 2x(4 - x)^{-1} + (2x - 5)(4 - x)^{-2}$ or $y' = \frac{2x}{4 - x} + \frac{2x - 5}{(4 - x)^2}$ looks wrong to me.

You've got your product rule okay, but when you differentiated $2x-5$ you got $2x$ rather than $2$.

A simple slip-up, it's so easy to do - other than that the technique is fine.

3. Originally Posted by Matt Westwood
$y' = 2x(4 - x)^{-1} + (2x - 5)(4 - x)^{-2}$ or $y' = \frac{2x}{4 - x} + \frac{2x - 5}{(4 - x)^2}$ looks wrong to me.

You've got your product rule okay, but when you differentiated $2x-5$ you got $2x$ rather than $2$.

A simple slip-up, it's so easy to do - other than that the technique is fine.
ah so I did. Thanks for the help.

4. Originally Posted by VitaX
Find the derivative of the following:

$y = (2x - 5)(4 - x)^{-1}$

I got $y' = 2x(4 - x)^{-1} + (2x - 5)(4 - x)^{-2}$ or $y' = \frac{2x}{4 - x} + \frac{2x - 5}{(4 - x)^2}$

Correct?
Personally, I'd use the quotient rule since $y = \frac{2x - 5}{4 - x}$.

5. Originally Posted by mr fantastic
Personally, I'd use the quotient rule since $y = \frac{2x - 5}{4 - x}$.
Yah works a lot better and easier here. Final answer is $y' = \frac{3}{(4 - x)^2}$