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Suppose $f$ and $g$ are smooth functions with $k+1$ derivatives and consider the composition $h=f\circ g.$ The chain rule implies that h is differentiable and that

$h'(t)=f'(g(t))g'(t)$

I didn't understand what is the meaning of last step?

Would anyone explain me the last step?

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    The last part isn't really a "step". It is telling you something without proof. $\frac{d}{dx}f(g(x)) = f'(g(x))*g'(x)$ – 2017-01-28

2 Answers 2

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In Leibniz notation, the last "step" can be seen as follows:

$$\frac{d\ f(g)}{dt}=\frac{d\ f(g)}{dg}\frac{dg}{dt}$$

Intuitively, you could imagine the "$dg$" cancelling to give you equality, but you'll have to take care of any divisions by $0$ and similar things. As per what it means, I think an example is good.

Let $f(t)=\sin(t)$ and $g(t)=t^2$.

then $f'(t)=\cos(t)$ and $g'(t)=2t$.

By the chain rule, we finally have

$$\frac d{dt}\sin(t^2)=\cos(t^2)\cdot2t$$

As I like to say, take the derivative of the outside $(f')$ and keep the inside $(g)$ the same, then multiply the result by the derivative of the inside $(g')$.

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    Beutiful Art, Does it mean $f$ is the functon of $g$ and $g$ is the function of $x$? – 2017-01-28
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    @DhamnekarV. That is what $f\circ g$ or $f(g(x))$ or $f(g)$ means. – 2017-01-28
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The chain rule says that $$h^\prime (t) = f^\prime (g(t)) g^\prime (t).$$ Example: Let $f(x) = x^2$ and $g(x) = 1-x$ (so $f(g(x)) = (1-x)^2$). Then, \begin{align} h^\prime (x) & = f^\prime ((1-x)) (1-x)^\prime \\ & = 2(1-x) \cdot (-1) \\ & = 2(x-1). \end{align}