# limit definition of the derivative problem

• Oct 16th 2008, 07:50 PM
VkL
limit definition of the derivative problem
Use algebra and the limit definition of the derivative to
find f′(−1), when

$\displaystyle f(x) = \frac {2}{1 -5x}$

I know the limit definition of the derivative is:
as the Lim h-->0 [F(x+a)-f(x)]/h
I get confused on the actual agebra part.
If anyone can kindly work it out, that would be very nice. thank you.
• Oct 16th 2008, 08:29 PM
Shyam
$\displaystyle f\left( x \right) = \frac{2} {{1 - 5x}} \hfill \\$

$\displaystyle f\left( {x + h} \right) = \frac{2} {{1 - 5\left( {x + h} \right)}} \hfill \\$

$\displaystyle f\left( {x + h} \right) - f\left( x \right) = \frac{2} {{1 - 5\left( {x + h} \right)}} - \frac{2} {{1 - 5x}} \hfill \\$

$\displaystyle = \frac{2} {{1 - 5x - 5h}} - \frac{2} {{1 - 5x}} \hfill \\$

$\displaystyle = \frac{{2\left( {1 - 5x} \right) - 2\left( {1 - 5x - 5h} \right)}} {{\left( {1 - 5x - 5h} \right)\left( {1 - 5x} \right)}} \hfill \\$

$\displaystyle = \frac{{2 - 10x - 2 + 10x + 10h}} {{\left( {1 - 5x - 5h} \right)\left( {1 - 5x} \right)}} \hfill \\$

$\displaystyle = \frac{{10h}} {{\left( {1 - 5x - 5h} \right)\left( {1 - 5x} \right)}} \hfill \\$

$\displaystyle {\text{Now,}} \hfill \\$

$\displaystyle f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}} {h} = \mathop {\lim }\limits_{h \to 0} \frac{{10h}} {{h\left( {1 - 5x - 5h} \right)\left( {1 - 5x} \right)}} \hfill \\$

$\displaystyle = \mathop {\lim }\limits_{h \to 0} \frac{{10}} {{\left( {1 - 5x - 5h} \right)\left( {1 - 5x} \right)}} = \frac{{10}} {{\left( {1 - 5x - 0} \right)\left( {1 - 5x} \right)}} \hfill \\$

$\displaystyle = \frac{{10}} {{\left( {1 - 5x} \right)\left( {1 - 5x} \right)}} = \frac{{10}} {{\left( {1 - 5x} \right)^2 }} \hfill \\$

$\displaystyle f'\left( x \right) = \frac{{10}} {{\left( {1 - 5x} \right)^2 }} \hfill \\$

$\displaystyle f'\left( { - 1} \right) = \frac{{10}} {{\left[ {1 - 5\left( { - 1} \right)} \right]^2 }} = \frac{{10}} {{36}} = \frac{5} {{18}} \hfill \\$
• Oct 16th 2008, 08:46 PM
11rdc11
Quote:

Originally Posted by VkL
Use algebra and the limit definition of the derivative to
find f′(−1), when

f(x) = 2/(1-5x)

I know the limit definition of the derivative is:
as the Lim h-->0 [F(x+a)-f(x)]/h
I get confused on the actual agebra part.
If anyone can kindly work it out, that would be very nice. thank you.

$\displaystyle \lim_{h \to 0}~\frac{\frac{2}{1-5(x+h)} - \frac{2}{1-5x}}{h}$

$\displaystyle \lim_{h \to 0}~\frac{\frac{2}{1-5x-5h} - \frac{2}{1-5x}}{h}$

$\displaystyle \lim_{h \to 0}~ \frac{\frac{2(1-5x) - 2(1-5x-5h)}{(1-5x-5h)(1-5x)}}{h}$

$\displaystyle \lim_{h \to 0}~\frac{\frac{2-10x-2+10x+10h}{1-5x-5h-5x+25x^2+25xh}}{h}$

$\displaystyle \lim_{h \to 0}~ \frac{\frac{10h}{1-10x+25x^2-5h+25xh}}{h}$

$\displaystyle \lim_{h \to 0}~ \frac{10h}{h(1-10x+25x^2-5h+25xh)}$

$\displaystyle \lim_{h \to 0}~ \frac{10}{1-10x+25x^2-5h+25xh}$

$\displaystyle \frac{10}{1-10x+25x^2}$
• Oct 17th 2008, 11:08 AM
VkL
which 1 is correct
which 1 is correct?
• Oct 17th 2008, 11:27 AM
ThePerfectHacker
Quote:

Originally Posted by VkL
which 1 is correct?

They are both correct.

Because $\displaystyle 1-10x+25x^2 = (1-5x)^2$.

Thus, $\displaystyle f'(x) = \frac{10}{(1-5x)^2}$.

Now let $\displaystyle x=1$ thus,
$\displaystyle f'(1) = \frac{10}{(1-5)^2} = \frac{10}{16}$.