# Thread: limit definition of the derivative problem

1. ## limit definition of the derivative problem

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

$
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.

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

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

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

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

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

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

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

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

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

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

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

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

$f'\left( { - 1} \right) = \frac{{10}}
{{\left[ {1 - 5\left( { - 1} \right)} \right]^2 }} = \frac{{10}}
{{36}} = \frac{5}
{{18}} \hfill \\$

3. 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.

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

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

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

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

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

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

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

$\frac{10}{1-10x+25x^2}$

4. ## which 1 is correct

which 1 is correct?

5. Originally Posted by VkL
which 1 is correct?
They are both correct.

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

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

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