# Intro Calc

• Jun 22nd 2007, 10:22 AM
johntuan
Intro Calc
Find the derivative using first principles

y= 3
sqrtx

Can someone help me solve this tks
• Jun 22nd 2007, 10:28 AM
Jhevon
Quote:

Originally Posted by johntuan
Find the derivative using first principles

y= 3
sqrtx

Can someone help me solve this tks

Differentiating using first principles means they want you to use the limit definition of a derivative. There are two equivalent definitions that are used, which one did you learn?

Definition 1:

For a function $f(x)$, it's derivative, $f'(x)$, at an arbitrary point $x$ is given by:

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

Defintion 2:

For a function $f(x)$, it's derivative, $f'(x)$, at an arbitrary point $a$ is given by:

$f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}$

I think the first definition is what you want, so just plug the pieces into the formula and find the limit. (Do you need help with that?)
• Jun 22nd 2007, 10:43 AM
johntuan
i learnt both ways but can you solve it using definition 1 please. And yes i need help, when i plug it in I keep getting stuck.
• Jun 22nd 2007, 10:58 AM
Jhevon
Quote:

Originally Posted by johntuan
i learnt both ways but can you solve it using definition 1 please

$f(x) = \frac {3}{ \sqrt {x}}$

$\Rightarrow f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h}$

$\Rightarrow f'(x) = \lim_{h \to 0} \frac { \frac {3}{ \sqrt {x + h}} - \frac {3}{ \sqrt {x}}}{h}$ ...........................................Combine the fractions in the numerator

$\Rightarrow f'(x) = \lim_{h \to 0} \frac {\frac {3 \sqrt {x} - 3 \sqrt {x + h}}{\sqrt {x} \sqrt {x + h}}}{h}$ ..............................................Now combined the fractions once again

$\Rightarrow f'(x) = \lim_{h \to 0} \frac {3 \sqrt {x} - 3 \sqrt {x + h}}{h \sqrt {x} \sqrt {x + h}}$ .....................................Now multiply by the conjugate of the numerator over itself (This is a standard approach).

$\Rightarrow f'(x) = \lim_{h \to 0} \frac {3 \sqrt {x} - 3 \sqrt {x + h}}{h \sqrt {x} \sqrt {x + h}} \cdot \frac {3 \sqrt {x} + 3 \sqrt {x + h}}{3 \sqrt {x} + 3 \sqrt {x + h}}$ ...........Now simplify this

$\Rightarrow f'(x) = \lim_{h \to 0} \frac {9x - 9(x + h)}{h \left( \sqrt {x} \sqrt {x + h} \right) \left( 3 \sqrt {x} + 3 \sqrt {x + h} \right)}$ ...........Simplify more

$\Rightarrow f'(x) = \lim_{h \to 0} \frac {-9h}{h \left( \sqrt {x} \sqrt {x + h} \right) \left( 3 \sqrt {x} + 3 \sqrt {x + h} \right)}$ ............Now we see that the $h$'s cancel

$\Rightarrow f'(x) = \lim_{h \to 0} \frac {-9}{ \left( \sqrt {x} \sqrt {x + h} \right) \left( 3 \sqrt {x} + 3 \sqrt {x + h} \right)}$ ...............Now there's no problem when we take the limit

$\Rightarrow f'(x) = \frac {-9}{6 \sqrt {x} x}$

$\Rightarrow f'(x) = - \frac {3}{2} x^{- \frac {3}{2}}$
• Jun 22nd 2007, 11:02 AM
johntuan
ok thanks i see where i messed up