# derivate

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• Nov 4th 2006, 01:30 AM
m777
derivate
please try to do these questions.
• Nov 4th 2006, 04:07 AM
topsquark
Quote:

Originally Posted by m777
please try to do these questions.

1. Find the derivative of $(2 \sqrt{x} + 2)(x - \sqrt{x})$.

This uses the product rule, so:
$\frac{d}{dx}[(2 \sqrt{x} + 2)(x - \sqrt{x})]$

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

= $\left ( \frac{1}{\sqrt{x}} \right )(x - \sqrt{x}) + (\sqrt{x} + 1) \left ( 2 - \frac{1}{\sqrt{x}} \right )$ (I've just moved around the 2 in the second term here.)

= $\frac{x}{\sqrt{x}} - 1 + 2\sqrt{x} - 1 + 2 - \frac{1}{\sqrt{x}}$

= $3\sqrt{x} - \frac{1}{\sqrt{x}}$

= $\frac{3x - 1}{\sqrt{x}}$

-Dan
• Nov 4th 2006, 06:35 AM
earboth
nr. 3 only
Quote:

Originally Posted by m777
please try to do these questions.

Hello, m777,

$f(x)=\left( \sqrt{x}-\frac{1}{\sqrt{x}} \right)^2$. Thus using chain rule

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

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

Expand the paranthese:

$f'(x)=1+\frac{1}{x}-\frac{1}{x}-\frac{1}{x^2}=1-\frac{1}{x^2}$

EB
• Nov 4th 2006, 09:08 AM
Soroban
Hello, m777!

Quote:

1) Prove that the derivative of . $y \:=\:\left(2\sqrt{x} + 2\right)\left(x - \sqrt{x}\right)$ .is: . $\frac{dy}{dx} \:=\:\frac{3x - 1}{\sqrt{x}}$

Multiply first: . $y \:=\:\left(2\sqrt{x} + 2\right)\left(x - \sqrt{x}\right) \;=\;2x\sqrt{x} - 2x + 2x - 2\sqrt{x}
$

And we have: . $y \;= \;2x^{\frac{3}{2}} - 2x^{\frac{1}{2}}$

Then: . $\frac{dy}{dx}\;=\;2\cdot\frac{3}{2}x^{\frac{1}{2}} - 2\cdot\frac{1}{2}x^{-\frac{1}{2}} \;= \;3x^{\frac{3}{2}} - x^{-\frac{1}{2}}$

Multiply "top and bottom" by $x^{\frac{1}{2}}:\;\;\frac{dy}{dx}\;= \;\frac{x^{\frac{1}{2}}}{x^{\frac{1}{2}}}\cdot\fra c{3x^{\frac{1}{2}} - x^{-\frac{1}{2}}}{1} \;= \;\frac{3x - 1}{x^{\frac{1}{2}}}$

Therefore: . $\boxed{\frac{dy}{dx} \;= \;\frac{3x - 1}{\sqrt{x}}}$

Quote:

3) Differentiate: . $y \:=\:\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2$

Square first: . $y \;= \;\left(\sqrt{x}\right)^2 - 2\left(\sqrt{x}\right)\left(\frac{1}{\sqrt{x}}\rig ht) + \left(\frac{1}{\sqrt{x}}\right)^2 \;= \;x - 2 + \frac{1}{x}$

So we have: . $y \;= \;x - 2 + x^{-1}$

Then: . $\frac{dy}{dx} \;= \;1 - x^{-2} \;= \;1 - \frac{1}{x^2}$

Therefore: . $\boxed{\frac{dy}{dx} \;= \;\frac{x^2 - 1}{x^2}}$

Quote:

2) Differentiate: . $y \;= \;\frac{\sqrt{a+x} + \sqrt{a - x}}{\sqrt{a+x} - \sqrt{a-x}}$

I'm still working on this one, but I would "rationalize" first.

Multiply top and bottom by $\left(\sqrt{a+x} + \sqrt{a-x}\right)$

. . $y \;= \;\frac{\sqrt{a+x} + \sqrt{a - x}}{\sqrt{a+x} + \sqrt{a - x}}\cdot\frac{\sqrt{a+x} + \sqrt{a+x}}{\sqrt{a-x} - \sqrt{a-x}} \;=\;\frac{\left(\sqrt{a+x} + \sqrt{a-x}\right)^2}{\left(\sqrt{a+x}\right)^2 - \left(\sqrt{a-x}\right)^2}$

. . $y \;= \;\frac{a + x + 2\sqrt{a+x}\sqrt{a-x} + a - x}{(a+x) - (a - x)} \;=\;\frac{2a + 2\sqrt{a^2-x^2}}{2x}$

. . $y \;= \;\frac{a + \sqrt{a^2-x^2}}{x}$ . . . then differentiate.

. . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I don't agree with the given answer.
. . (Why aren't the 2's cancelled out?)

Could it be: . $\frac{dy}{dx}\;= \;\frac{a}{a - \sqrt{a^2-x^2}}$ ?

• Nov 5th 2006, 03:38 AM
m777
answer
hello,
soroban
In my book answer is same as i mention in question.any how i will concern with my teacher but i agree with your answer.
many best wishes for you
from
m777.