Differentiate

**Awsom Guy** · February 12th 2010, 07:25 PM

Differentiate this function f(x)= (4x)/(sqrt2x).
I know I use the quotient rule to differentiate this, this is what I have done so far:

(2x)^1/2 (0) - (4) [1/2(2x)^-1/2] divided by (2x)

0 - 2(2x)^-1/2 divided by (2x)

-(2x)^-1/2 divided by x

-2 divided by x^1/2 * x

-2 divided by sqrtx^3

Please help if something here is confusing because i dont know how to use Latex please tell I will try to show it more better.
Thanks

**Nacho** · February 12th 2010, 07:37 PM

fastter than that, is note that: $f(x)=\dfrac{4}{\sqrt{2}}\cdot \dfrac{x}{\sqrt{x}}$

**Awsom Guy** · February 12th 2010, 07:46 PM

but what I have done is not right. Tha answer given is (2)/(sqrt2x)

**Keithfert488** · February 12th 2010, 07:51 PM

Originally Posted by Nacho

fastter than that, is note that: $f(x)=\dfrac{4}{\sqrt{2}}\cdot \dfrac{x}{\sqrt{x}}$

Nacho is pointing out that his approach and hint will work better. Instead, split it into a constant ( $\frac{4}{\sqrt 2}$ ) multiplied by a term in x ( $\frac{x}{\sqrt x}=\sqrt x$ )

So then you can just use the power rule. So before differentiating you have changed the function to $y=\frac{4\sqrt x}{\sqrt2}$

Can you differentiate that more easily?

**Nacho** · February 12th 2010, 07:56 PM

ok, I don´t undestand why you multiplicate by zero in the first line, if the quotient rule says: $ \left( {\frac{h} {g}} \right)^\prime = \frac{{h'g - hg'}} {{g^2 }} $

In your example $h=4x$ and $g=\sqrt{2x}$

Whit my way is easy $ f(x) = \frac{4} {{\sqrt 2 }} \cdot x^{1/2} \Rightarrow f'(x) = \frac{4} {{\sqrt 2 }} \cdot \frac{1} {2} \cdot x^{ - 1/2} = \frac{{\sqrt 2 }} {{\sqrt x }} $

**Keithfert488** · February 12th 2010, 08:02 PM

Originally Posted by Nacho

ok, I don´t undestand why you multiplicate by zero in the first line, if the quotient rule says: $ \left( {\frac{h} {g}} \right)^\prime = \frac{{h'g - hg'}} {{g^2 }} $

In your example $h=4x$ and $g=\sqrt{2x}$

Whit my way is easy $ f(x) = \frac{4} {{\sqrt 2 }} \cdot x^{1/2} \Rightarrow f'(x) = \frac{4} {{\sqrt 2 }} \cdot \frac{1} {2} \cdot x^{ - 1/2} = \frac{{\sqrt 2 }} {{\sqrt x }} $

Just pointing out simplification

$\frac{4}{\sqrt{2}}\cdot \frac{1}{2}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2}}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2x}}$

**Nacho** · February 12th 2010, 08:06 PM

Originally Posted by Keithfert488

Just pointing out an error in simplification

$\frac{4}{\sqrt{2}}\cdot \frac{1}{2}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2}}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2x}}$

I just continue the simplification

$ \frac{2} {{\sqrt 2 }} = \frac{{\sqrt 2 \sqrt 2 }} {{\sqrt 2 }} = \sqrt 2 $

**drumist** · February 12th 2010, 08:07 PM

Originally Posted by Keithfert488

Just pointing out an error in simplification

$\frac{4}{\sqrt{2}}\cdot \frac{1}{2}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2}}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2x}}$

Your solution and his solution are equal.

**Keithfert488** · February 12th 2010, 08:16 PM

Originally Posted by drumist

Your solution and his solution are equal.

Wow. I feel stupid. Well...I was doing it so we can see it through to the answer that he gave us.

**Awsom Guy** · February 12th 2010, 08:47 PM

lols thanks guys.

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