Induction Question (1 Viewer)

rand_althor · Jun 7, 2015

Hey, was not sure if you're meant to use the power rule for part (ii).

InteGrand · Jun 7, 2015

rand_althor said:
Hey, was not sure if you're meant to use the power rule for this question.

$(i) $\frac{\mathrm{d}}{\mathrm{d}x}(x)=\lim _{h\to 0}\frac{(x+h) - x}{h}=\lim _{h\to 0}\frac{h}{h}=\lim _{h\to 0}1 = 1$$

$(ii) The case for $n=1$ was proved in part (i). Suppose $\frac{\mathrm{d}}{\mathrm{d}x}(x^k)=kx^{k-1}$. We need to show that $\frac{\mathrm{d}}{\mathrm{d}x}(x^{k+1})=(k+1)x^k$. By the product rule, $\frac{ \mathrm{d}}{\mathrm{d}x}(x^{k+1})=\frac{\mathrm{d}}{\mathrm{d}x}(x\cdot x^k)=\frac{\mathrm{d}}{\mathrm{d}x}(x) \cdot x^k + x\cdot \frac{ \mathrm{d}}{\mathrm{d}x}(x^k) = x^k + x\cdot kx^{k-1}$ (using the inductive hypothesis and that $\frac{\mathrm{d}}{\mathrm{d}x}(x) = 1$) $= x^k + kx^k = (k+1)x^k$, as required. Hence proved by induction. \blacksquare$

$We can't use the power rule, since that is precisely what we're trying to prove.$

Induction Question (1 Viewer)

rand_althor

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InteGrand

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