1. ## [SOLVED] Differentiation

I have been trying to do this question for like a week, and I have had no success.
Question:
Differentiate the function f(x)=(4x)/(sqrt2x)
I have tried it and I know that I have to use the quotient rule but I know I am doing something wrong in this step.

(2x)^1/2 * (4) - (4x) * (x)^-1/2 all divided by (2x)^3/2

2. Originally Posted by Awsom Guy
I have been trying to do this question for like a week, and I have had no success.
Question:
Differentiate the function f(x)=(4x)/(sqrt2x)
I have tried it and I know that I have to use the quotient rule but I know I am doing something wrong in this step.

(2x)^1/2 * (4) - (4x) * (x)^-1/2 all divided by (2x)^3/2

$f(x) = \frac{4x}{\sqrt{2x}}$

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

Now using the product rule:

$f'(x) = 4x\frac{d}{dx}[(2x)^{-\frac{1}{2}}] + (2x)^{-\frac{1}{2}}\frac{d}{dx}(4x)$

$= 4x\cdot 2\cdot \left(-\frac{1}{2}(2x)^{-\frac{3}{2}}\right) + 4(2x)^{-\frac{1}{2}}$

$= 4(2x)^{-\frac{1}{2}} - 4x(2x)^{-\frac{3}{2}}$

$= \frac{4}{\sqrt{2x}} - \frac{4x}{(\sqrt{2x})^3}$

$= \frac{4}{\sqrt{2x}}\left[1 - \frac{x}{(\sqrt{2x})^2}\right]$

$= \frac{4}{\sqrt{2x}}\left(1 - \frac{x}{2x}\right)$

$= \frac{4}{\sqrt{2x}}\left(1 - \frac{1}{2}\right)$

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

3. ## helps

Originally Posted by Prove It
$f(x) = \frac{4x}{\sqrt{2x}}$

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

Now using the product rule:

$f'(x) = 4x\frac{d}{dx}[(2x)^{-\frac{1}{2}}] + (2x)^{-\frac{1}{2}}\frac{d}{dx}(4x)$

$= 4x\cdot 2\cdot \left(-\frac{1}{2}(2x)^{-\frac{3}{2}}\right) + 4(2x)^{-\frac{1}{2}}$

$= 4(2x)^{-\frac{1}{2}} - 4x(2x)^{-\frac{3}{2}}$

$= \frac{4}{\sqrt{2x}} - \frac{4x}{(\sqrt{2x})^3}$

$= \frac{4}{\sqrt{2x}}\left[1 - \frac{x}{(\sqrt{2x})^2}\right]$ummm.... how did the power go from 3 to 2.

$= \frac{4}{\sqrt{2x}}\left(1 - \frac{x}{2x}\right)$

$= \frac{4}{\sqrt{2x}}\left(1 - \frac{1}{2}\right)$

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

4. Originally Posted by Awsom Guy
I took out $\frac{4}{\sqrt{2x}}$ as a common factor.

Note that $(\sqrt{2x})^3 = \sqrt{2x}(\sqrt{2x})^2$.

5. Oh ok thanks for that. Note: I have been stuck with that question for almost a week. Thanks

6. Like Prove It, I recommend using negative powers and the product rule rather than positive powers and the quotient rule. However, since you specifically asked about that:

(2x)^1/2 * (4) - (4x) * (x)^-1/2 all divided by (2x)^3/2

$\frac{(2x)^{1/2}(4)- (4x)(x)^{-1/2}}{(2x)^{3/2}}$

You have a typo in the second term of the numerator: it should be [tex](2x)^{-1/2}[tex], not $(x)^{-1/2}$. Also the denominator is wrong. In the quotient rule, the quotient of the derivative is the square of the denominator of the function. Here, that is $((2x)^{1/2})^2= 2x$. You should have
$\frac{(2x)^{1/2}(4)- (4x)(2x)^{-1/2}}{2x}$

To simplify that, factor $(2x)^{-1/2}$ out of the numerator.

$(2x)^{1/2}= \frac{(2x)}{(2x)^{-1/2}}$ so your numerator becomes $(2x)^{-1/2}[(2x)(4)- 4x]= (2x)^{-1/2}(4x)$.

Putting that into half power into the denominator makes it $(2x)^{1/2}$ and, of course, $(2x)^{1/2}(2x)= (2x)^{3/2}$.

You have a 3/2 power in the denominator after simplifying:
$\frac{4x}{(2x)^{3/2}}= \frac{4x}{\left(\sqrt{2x}\right)^3}$

(Since the original problem was given in terms of the square root rather than a fractional power, it is probably better to give the answer that way.)

7. umm..lols thanks I have already solved this :0 thanks anyways Ill close this thread