First Principal derivatives

**pr3m** · July 24th 2009, 12:31 PM

Hello, im new here but im doing a calculus course online a grade 12 University course and im stuck on how to do first principals derivatives, and since i have no teacher just learning off the book its tough if you guys dont mind help me solve these questions for a practice so i can start to hand in work.

a) f(x)=x²-2x+1
b) f(x)=x³-3x²
c) f(x)=(√2x)
d) f(x)=4/x+1

greatly appreciated

**craig** · July 24th 2009, 01:48 PM

Originally Posted by pr3m

Hello, im new here but im doing a calculus course online a grade 12 University course and im stuck on how to do first principals derivatives, and since i have no teacher just learning off the book its tough if you guys dont mind help me solve these questions for a practice so i can start to hand in work.

a) f(x)=x²-2x+1
b) f(x)=x³-3x²
c) f(x)=(√2x)
d) f(x)=4/x+1

greatly appreciated

If you are doing calculus then you should be aware of the basics, that is that the derivative of $x^n$ is $nx^{n-1}$ , where n is a constant.

For your first one, you have $f(x)=x^2-2x+1$ , $f'(x)=2x-2$ .

b) $f(x)=x^3-3x^2$ - follow the steps above.

c) $f(x)=\sqrt{2x}$ , this can be written as $2x^{\frac{1}{2}}$ , follow the steps above.

d) $f(x)= \frac{4}{x+1}$ , this can be written as $4(x+1)^{-1}$ , follow the steps above.

If there is anything you are unsure about just ask =)

**pr3m** · July 24th 2009, 01:58 PM

hey thanks for the response... i know how to find the derivative of somthing to solve for f prime its just with first principals they put it in the other form where its like f(x+h)-f(x) / h =... thats what im having trouble with finding the steps to solve each one and geting it where u go lim h -> 0 or w\e know what i mean?

**yeongil** · July 24th 2009, 04:08 PM

Are you asking to find the derivatives using the limit definition? (BTW, this should have been posted in the Calculus subforum.)

Then, using this:
$\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

For (a), we get
$\lim_{h \to 0} \frac{[(x + h)^2 - 2(x + h) + 1] - [x^2 - 2x + 1]}{h}$

$= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - 2x - 2h + 1 - x^2 + 2x - 1}{h}$

$= \lim_{h \to 0} \frac{2xh + h^2 - 2h}{h}$

$= \lim_{h \to 0} (2x + h - 2)$

As $h \rightarrow 0$ ,
$\lim_{h \to 0} (2x + h - 2) = 2x - 2$ .

01

**Deco** · July 24th 2009, 05:37 PM

Generally, first derivatives for the ones you listed above, involving powers
and polynomials; $x^n \rightarrow nx^{n-1}$

This is a rather sloppy and shorthand way of showing what has
been demonstrated above.

(The $\longrightarrow$ represents the process of differentiating)

So,

(1) $x^5 \longrightarrow 5x^4$

(2) $x^1 \longrightarrow 1x^0 = 1; x \neq 0$

(3) $4x^3 + 5 = 4x^3 +5x^0 \longrightarrow 12x^2 + 0 = 12x^2$

**earboth** · July 24th 2009, 11:03 PM

Originally Posted by pr3m

... im stuck on how to do first principals derivatives,
...

c) f(x)=(√2x)

...

I'll do this step by step:

$f'(x)=\lim_{ h \to 0}\dfrac{\sqrt{2(x+h)} - \sqrt{2x}}{h}$

${\color{white}f'(x)}=\lim_{ h \to 0}\dfrac{\sqrt{2(x+h)} - \sqrt{2x}}{h} \cdot \dfrac{\sqrt{2(x+h)} + \sqrt{2x}}{\sqrt{2(x+h)} + \sqrt{2x}}$

${\color{white}f'(x)}=\lim_{ h \to 0}\dfrac{2x+2h-2x}{h \left(\sqrt{2(x+h)} + \sqrt{2x} \right)}$

${\color{white}f'(x)}=\lim_{ h \to 0}\dfrac{2}{\sqrt{2(x+h)} + \sqrt{2x} }$

.................................................. .......... Since $\lim_{h \to 0} \sqrt{2(x+h)} = \sqrt{2x}$ the last limit simplifies to:

${\color{white}f'(x)}= \dfrac{2}{\sqrt{2x} + \sqrt{2x} }= \dfrac{2}{2 \cdot \sqrt{2x} }$
Thus:
$\boxed{f'(x)=\dfrac1{\sqrt{2x}}}$

Thread: First Principal derivatives

LinkBack

Thread Tools

Display

First Principal derivatives

Similar Math Help Forum Discussions

Principal Ideal

Product of non principal ideals is principal.

principal value mapping

Cauchy Principal value

Principal Ideal

Search Tags