Question
Find the derivative of (x3 - 27) from first principle.
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Answer
We have, f(x) = x3 - 27
By using first principle of derivative,
${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$
$\therefore \quad {f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{(x + h)}^3} - 27} \right] - \left[ {{x^3} - 27} \right]}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{{x^3} + 3{x^2}h + 3{h^2}x + {h^3} - 27 - {x^3} + 27}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{3{x^2}h + 3x{h^2} + {h^3}}}{h}$
$ = \mathop {\lim }\limits_{h \to 0}$ ( 3x2 + 3xh + h2) = 3x2
By using first principle of derivative,
${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$
$\therefore \quad {f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{(x + h)}^3} - 27} \right] - \left[ {{x^3} - 27} \right]}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{{x^3} + 3{x^2}h + 3{h^2}x + {h^3} - 27 - {x^3} + 27}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{3{x^2}h + 3x{h^2} + {h^3}}}{h}$
$ = \mathop {\lim }\limits_{h \to 0}$ ( 3x2 + 3xh + h2) = 3x2
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