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Suppose I have the function $f(x)=x^3$. The derivative is obviously $f'(x) = 3x^2$. But $3x^2$ is nonlinear since $$f'(3x) = 27x^2$$ $$3f'(x) = 9x^2$$ Therefore this isn't a linear map.

Rudin defines the following $$f(x+h)-f(x) = f'(x)h + r(h)$$ where $r(h)$ is very small, and he says we can regard the derivative of $f$ at $x$ as the linear operator that maps $$h\mapsto f'(x)h$$

How does this make sense?

Stan Shunpike
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    As you wrote, $f'$ is not linear but $h\mapsto f'(x)h$ is. The mistake you made is that you consider $x \mapsto f'(x)$ and not $h\mapsto f'(x)h$. – Abelmondo Jul 10 '15 at 00:20

2 Answers2

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$h\mapsto f'(x)h$ is a linear map of $\boldsymbol{h}$, not of $x$, which is only a fixed parameter in such questions. Isn't the tangent a straight line anyway?

Bernard
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  • So for any two functions $f$ and $g$, $(f+g)' = f' + g'$ and for any scalar $c$, $(cf)' = c(f')$. – Neal Lawton Jul 10 '15 at 01:24
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    @NealLawton That shows that differentiation is a linear map. This question is asking about how the derivative is a linear map (it gives the best linear approximation to the rate of change at a point). –  Jul 10 '15 at 01:35
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It is linear at each point. For example $f'(2)=12$. So, the linear map this induces is $x\mapsto 12x$.

J126
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