# Math Help - First Principal derivatives

1. ## First Principal derivatives

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

2. 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 =)

3. 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?

4. 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

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

6. 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:
$\boxed{f'(x)=\dfrac1{\sqrt{2x}}}$