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
Thank you in advance.
First, recall that $\displaystyle f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$.
Now, from the given $\displaystyle f\left( x \right) = \frac{2}{{{x^3}}}$, you'll get $\displaystyle f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2{x^3} - 2{{\left( {x + h} \right)}^3}}}{{h{x^3}{{\left( {x + h} \right)}^3}}}$.
Whan you expand $\displaystyle \frac{{2{x^3} - 2{{\left( {x + h} \right)}^3}}}{{h{x^3}{{\left( {x + h} \right)}^3}}}$, you'll get $\displaystyle \frac{{2{x^3} - 2{x^3} - 6x{h^2} - 6{x^2}h - 2{h^3}}}{{{x^6}h + 3{x^4}{h^3} + 3{x^5}{h^2} + {x^3}{h^4}}}$.
We can cancel an $\displaystyle h$ from each of the terms in the numerator and denominator and we know that $\displaystyle 2x^3 - 2x^3 = 0$ and thus we have:
$\displaystyle f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{ - 6xh - 6{x^2} - 2{h^2}}}{{{x^6} + 3{x^4}{h^2} + 3{x^5}h + {x^3}{h^3}}}$
Since $\displaystyle h$ tends to zero, each of the terms with an $\displaystyle h$ will be zero. You'll be left with $\displaystyle f'\left( x \right) = \frac{{ - 6{x^2}}}{{{x^6}}} = - 6{x^{ - 4}}$.
You made a mistake in expanding $\displaystyle \frac{{2{x^3} - 2{{\left( {x + h} \right)}^3}}}{{h{x^3}{{\left( {x + h} \right)}^3}}}$. This is not equal to $\displaystyle \frac{{2{x^3} - 2{x^3} - 2{h^3}}}{{h{{\left( {x + h} \right)}^3}{x^3}}}$ because $\displaystyle - 2{\left( {x + h} \right)^3} \ne - 2{x^3} - 2{h^3}$.
I hope this helps you.
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