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differentiation from first principles

differentiation from first principles

Given a curve, with equation $y = f(x)$, the gradient of the curve at the point $P(x_0,y_0)$ is given by the limit $$ f'(x_0) = \lim_{h\rightarrow 0}\left(\frac{f(x_0+h) - f(x_0)}{h}\right) $$

(Technical caveat - where the limit exists)

differentiation from first principles 1

Determine the gradient of the curve $y = x^3$ at the point $(2,8)$
solution - press button to display

From the definition of the derivative at the point $(x_0,y_0)$, we get $$ f'(x_0) = \lim_{h\rightarrow 0}\frac{(x_0+h)^3 - x_0^3) }{h} $$ Taking $x_0 = 2$, gives $$ f'(2) = \lim_{h\rightarrow 0}\frac{(2+h)^3 - 2^3) }{h} $$ Now applying the binomial expansion to the top of the fraction yields $$ f'(2) = \lim_{h\rightarrow 0}\frac{8 + 12h + 6h^2 + h^3 - 8 }{h} $$ Cancelling the constant terms gives $$ f'(2) = \lim_{h\rightarrow 0}\frac{ 12h + 6h^2 + h^3 }{h} $$ Diving by $h$ $$ f'(2) = \lim_{h\rightarrow 0}(12 + 6h + h^2) $$ As $h\rightarrow 0$, $6h\rightarrow 0$, $h^2\rightarrow 0$ and hence $f'(2)\rightarrow 12$

We conclude that the gradient of the curve $y = x^3$ at $(2,8)$ is $12$.