# Math Help - Definition of derivative

1. ## Definition of derivative

Please I need the steps for the solution.

by using definition of derivative find the f'(x)

the question is

f(x)= 2/x^3

2. Originally Posted by Abdul
Please I need the steps for the solution.

by using definition of derivative find the f'(x)

the question is

f(x)= 2/x^3

From first principles:

$f'(x) = \frac {\frac 2 {(x+h)^3} - \frac 2{x^3}} h$

Expand that out and simplify, then let h tend to zero.

3. I could not get the final answer correct.

see what I did

$f'(x)= \frac {2x^3 -2x^3 -2h^3} {(h)(x+h)^3(x^3)}$

Then I got

$f'(x)= \frac {-2h^3} {(h)(x+h)^3(x^3)}$

What to do next?

4. First, recall that $f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$.

Now, from the given $f\left( x \right) = \frac{2}{{{x^3}}}$, you'll get $f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2{x^3} - 2{{\left( {x + h} \right)}^3}}}{{h{x^3}{{\left( {x + h} \right)}^3}}}$.

Whan you expand $\frac{{2{x^3} - 2{{\left( {x + h} \right)}^3}}}{{h{x^3}{{\left( {x + h} \right)}^3}}}$, you'll get $\frac{{2{x^3} - 2{x^3} - 6x{h^2} - 6{x^2}h - 2{h^3}}}{{{x^6}h + 3{x^4}{h^3} + 3{x^5}{h^2} + {x^3}{h^4}}}$.

We can cancel an $h$ from each of the terms in the numerator and denominator and we know that $2x^3 - 2x^3 = 0$ and thus we have:

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{ - 6xh - 6{x^2} - 2{h^2}}}{{{x^6} + 3{x^4}{h^2} + 3{x^5}h + {x^3}{h^3}}}$

Since $h$ tends to zero, each of the terms with an $h$ will be zero. You'll be left with $f'\left( x \right) = \frac{{ - 6{x^2}}}{{{x^6}}} = - 6{x^{ - 4}}$.

You made a mistake in expanding $\frac{{2{x^3} - 2{{\left( {x + h} \right)}^3}}}{{h{x^3}{{\left( {x + h} \right)}^3}}}$. This is not equal to $\frac{{2{x^3} - 2{x^3} - 2{h^3}}}{{h{{\left( {x + h} \right)}^3}{x^3}}}$ because $- 2{\left( {x + h} \right)^3} \ne - 2{x^3} - 2{h^3}$.

I hope this helps you.

5. Thanks a lot for your help. now I got where I made a mistake.

6. Originally Posted by Abdul
Thanks a lot for your help. now I got where I made a mistake.
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