A few Questions on the derivative

**walk_in_fark** · June 4th 2009, 10:14 PM

Hello, I had a few questions that I completed and I'm unsure if my answers are correct. i had to find the derivative of the √2x, using first principle, f(a+h) - f(a)/h. The answer that I arrived at is 1/√2√x, but when i use the power rule i get 1/x^2. Are they both wrong, or am i doing something wrong, somewhere? The second question is finding the derivative of 3x/(x^2 + 4), the answer i arrived at, is 3/x^2 + 4 - 6x^2/(x^2 + 4)^2. I got this using the quotient rule. Sorry about the mess, i'm completly new to calculus, have no one around me to check my answers until monday and, I just want to know my mistakes. Again, thanks very much.

**Rapha** · June 4th 2009, 11:44 PM

Hi.

Originally Posted by walk_in_fark

Hello, I had a few questions that I completed and I'm unsure if my answers are correct. i had to find the derivative of the √2x, using first principle, f(a+h) - f(a)/h. The answer that I arrived at is 1/√2√x, but when i use the power rule i get 1/x^2. Are they both wrong, or am i doing something wrong, somewhere?

No, the first solution is correct, I guess you mean $f'(x) = \frac{1}{\sqrt 2 \sqrt x}$ , don't you?

$f(x) = \sqrt{2x} = \sqrt{2} \sqrt{x} = \sqrt{2} x^{1/2}$

Thus
$f'(x) = \sqrt 2 \frac 1 2 x^{-1/2} =\frac{ \sqrt 2}{2} * \frac{1}{x^{0.5}}$

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

Originally Posted by walk_in_fark

The second question is finding the derivative of 3x/(x^2 + 4), the answer i arrived at, is 3/x^2 + 4 - 6x^2/(x^2 + 4)^2. I got this using the quotient rule. Sorry about the mess, i'm completly new to calculus, have no one around me to check my answers until monday and, I just want to know my mistakes. Again, thanks very much.

Yes, that is correct, but I prefer

$f(x) = \frac{3x}{x^2+4}$

$f'(x) = \frac{3(4 - x^2)}{(x^2 + 4)^2}$ (this is the same solution...)

kind regards
Rapha

**Amer** · June 4th 2009, 11:44 PM

Originally Posted by walk_in_fark

Hello, I had a few questions that I completed and I'm unsure if my answers are correct. i had to find the derivative of the √2x, using first principle, f(a+h) - f(a)/h. The answer that I arrived at is 1/√2√x, but when i use the power rule i get 1/x^2. Are they both wrong, or am i doing something wrong, somewhere? The second question is finding the derivative of 3x/(x^2 + 4), the answer i arrived at, is 3/x^2 + 4 - 6x^2/(x^2 + 4)^2. I got this using the quotient rule. Sorry about the mess, i'm completly new to calculus, have no one around me to check my answers until monday and, I just want to know my mistakes. Again, thanks very much.

first one
I prefer to write it like

$\sqrt{2x}=(2x)^{\frac{1}{2}}$

$\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}$

$\lim_{h\rightarrow0}\frac{\sqrt{x+h}-\sqrt{x}}{h}=\lim_{h\rightarrow0}\left(\frac{\sqrt {x+h}-\sqrt{x}}{h}\right)\left(\frac{\sqrt{x+h}+\sqrt{x} }{\sqrt{x+h}+\sqrt{x}}\right)$

$\lim_{h\rightarrow0}\left(\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\right)=\lim_{h\rightarr ow0}\left(\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\right)$

the rest for you just sub zero instead of h after delete h from the denominator and the numentor

**Amer** · June 4th 2009, 11:50 PM

Originally Posted by walk_in_fark

3x/(x^2 + 4), the answer i arrived at, is 3/x^2 + 4 - 6x^2/(x^2 + 4)^2. .

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

use product rule

$H(x)=g(x)f(x)....H'(x)=g'(x)f(x)+f'(x)g(x)$

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