1. ## Differentiate

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

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

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

4. Originally Posted by Nacho
fastter than that, is note that: $\displaystyle f(x)=\dfrac{4}{\sqrt{2}}\cdot \dfrac{x}{\sqrt{x}}$
Nacho is pointing out that his approach and hint will work better. Instead, split it into a constant ($\displaystyle \frac{4}{\sqrt 2}$) multiplied by a term in x ($\displaystyle \frac{x}{\sqrt x}=\sqrt x$)

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

Can you differentiate that more easily?

5. ok, I donīt undestand why you multiplicate by zero in the first line, if the quotient rule says: $\displaystyle \left( {\frac{h} {g}} \right)^\prime = \frac{{h'g - hg'}} {{g^2 }}$

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

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

6. Originally Posted by Nacho
ok, I donīt undestand why you multiplicate by zero in the first line, if the quotient rule says: $\displaystyle \left( {\frac{h} {g}} \right)^\prime = \frac{{h'g - hg'}} {{g^2 }}$

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

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

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

7. Originally Posted by Keithfert488
Just pointing out an error in simplification

$\displaystyle \frac{4}{\sqrt{2}}\cdot \frac{1}{2}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2}}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2x}}$
I just continue the simplification

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

8. Originally Posted by Keithfert488
Just pointing out an error in simplification

$\displaystyle \frac{4}{\sqrt{2}}\cdot \frac{1}{2}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2}}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2x}}$
Your solution and his solution are equal.

9. 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.

10. lols thanks guys.