1. Strange derivative question

We are given that:

$\displaystyle f(x)=\displaystyle{\frac{(x-4)^2}{2x^\frac{1}{2}}}$

Show that:

$\displaystyle f'(x)=\displaystyle{\frac{3x^2-8x-16}{4x^\frac{3}{2}}}$

I've tried to figure this out and it just won't work!

2. Hello, anthmoo!

It's not "strange" . . . You're making an error or two.

Given: .$\displaystyle f(x)\:=\:\frac{(x-4)^2}{2x^{\frac{1}{2}}}$

Show that: .$\displaystyle f'(x)\:=\:\frac{3x^2-8x-16}{4x^{\frac{3}{2}}}$

Quotient Rule: .$\displaystyle f'(x)\:=\:\frac{2x^{\frac{1}{2}}\cdot2(x-4) - (x-4)^2\cdot x^{-\frac{1}{2}}}{4x} \;=\;\frac{4x^{\frac{1}{2}}(x-4) - x^{-\frac{1}{2}}(x-4)^2}{4x}$

Multiply top and bottom by $\displaystyle x^{\frac{1}{2}}$

. . $\displaystyle \frac{x^{\frac{1}{2}}}{x^{\frac{1}{2}}}\cdot\frac{ 4x^{\frac{1}{2}}(x-4) - x^{-\frac{1}{2}}(x-4)^2}{4x}$ $\displaystyle = \;\frac{4x(x-4) - (x-4)^2}{4x^{\frac{3}{2}}} \;=\;\frac{4x^2 - 16x - (x^2 - 8x + 16)}{4x^{\frac{3}{2}}}$

Therefore: .$\displaystyle f'(x)\;=\;\frac{3x^2 - 8x - 16}{4x^{\frac{3}{2}}}$

3. Originally Posted by anthmoo
We are given that:

$\displaystyle f(x)=\displaystyle{\frac{(x-4)^2}{2x^\frac{1}{2}}}$

Show that:

$\displaystyle f'(x)=\displaystyle{\frac{3x^2-8x-16}{4x^\frac{3}{2}}}$

I've tried to figure this out and it just won't work!
Hello Anthmoo,

use quotient rule:

$\displaystyle f'(x)=\displaystyle{\frac{2x^\frac{1}{2}\cdot 2(x-4)-(x-4)^2\cdot 2 \cdot \frac{1}{2}x^{-\frac{1}{2}}}{4x}}$ = $\displaystyle \displaystyle{\frac{2x^\frac{1}{2}\cdot 2(x-4)-(x-4)^2\cdot 2 \cdot \frac{1}{2}\cdot \left({\frac{1}{\sqrt{x}}}\right)}{4x}}$

The common denominator in the numerator is $\displaystyle sqrt{x}$. You'll get:
$\displaystyle \displaystyle{\frac{\frac{4x(x-4)-(x-4)^2}{\sqrt{x}}}{4x}}$Expand the numerator and put together both denominators:

$\displaystyle \displaystyle{\frac{4x^2-16x-x^2+8x-16}{4x \cdot \sqrt{x}}}$ which is exactly the given result

EB

4. Thanks again guys!

Looking at yours I realised I did make a stupid error of originally not expanding correctly.

In an earlier part of the question, you were to expand without the denominator - there, 2x in the denominator instead of 4x!

Thanks!