1. ## Applying chain rule

Find the derivative of the following:

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

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

Correct?

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

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

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

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

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

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

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

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

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:

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

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

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

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