This is a problem that has been bugging me for ages. I just can't wrap my head around this weird result. I know I went wrong somewhere [as a matter of fact, that was the answer I was hoping for], but most sources, (including, but not limited to, wikipedia), suggest otherwise.(adsbygoogle = window.adsbygoogle || []).push({});

I will cut to the chase.

A function f is defined by $$f(x)=x^2$$

The derivative, f', is then $$f'(x)=2x$$

So far so good.

Here is where all hell broke loose for me:

The composition of f and f' yields $$f[f'(x)]=(2x)^2=4x^2$$

I will now switch to Leibniz Notation for convinience.

u=f(x), and v=f'(x)

According to the chain rule

[itex]\frac{du}{dx}[/itex]=[itex]\frac{du}{dv}[/itex][itex]\frac{dv}{dx}[/itex]

$$f'(x)=f'[f'(x)]f''(x)$$

Well, we know what f'(x) is, it is 2x as shown above.

du/dv is (2)(2x)=4x

And dv/dx is simply 2.

Putting all of this in the above equation, for nonzero x: $$2x=8x$$ $$2=8$$

My speculation as to why this bizarre result popped out was the fact that my substitution of f'(x) for x in f(x) to get f[f'(x)] was wrong. But after going through several websites, I came to realize that there is nothing mathematically wrong with this substitution.

Can someone please point out to me where I went wrong with this?

I would be more than grateful.

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# The Chain Rule and Function Composition

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