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Note: The inverse of $f(x)$, $f^{-1}(f(x)) = x$.

My interpretation of this: $1$ is equal to the derivative of $x$ which is equal to the derivative of the inverse function.

Then this next part is where it gets foggy for me...

The derivative of the inverse with respect to x is the function times its derivative...? But I just said what the derivative of the inverse was before...? Why is chain rule used here?

I assume the first part bottom line is because the functions derivative is $1$ but why does that become $\frac{1}{f'(x)}?$

If anyone's wondering, I am using this to prove the derivative of arcsin, but I would like as little help as possible, I just want to understand the problem is all, not a full solve. Thanks.

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I've just updated my answer to show how the Leibniz notation makes it a bit simpler. –  Michael Hardy Feb 22 at 17:06

4 Answers 4

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My interpretation of this: $1$ is equal to the derivative of $x$ which is equal to the derivative of the inverse function.

That is incorrect. $$\frac{d}{dx}f^{-1}(f(x))$$ is the derivative of the function $f^{-1}$ applied to the function $f$. For comparison, the derivative of the inverse function is $$\frac{d}{dx}f^{-1}(x).$$ Note that a different way of writing would be $$\frac{d}{dx}f^{-1}(f(x))=\frac{d}{dx}(f^{-1}\circ f)(x),$$ where $\circ$ denotes the composition of two functions. The fact that we have two functions being composed means that we should use the chain rule (see Wikipedia), which is precisely what occurs in the rest of the equations you are asking about.


The main confusion here is that the juxtaposition of two symbols can mean two different things: multiplication of two functions, and application of a function (or operation) to an input.

In the expression $$\frac{d}{dx}x$$ we apply the operation $\frac{d}{dx}$ to the function $x\mapsto x$ to produce the (constant) function $x\mapsto 1$.

In the expression $$\frac{d}{dx}f^{-1}(f(x))$$ we apply the operation $\frac{d}{dx}$ to the function $x\mapsto f^{-1}(f(x))$. Since $f^{-1}(f(x))=x$ for every $x$ (by the very definition of $f^{-1}$ being the inverse of $f$), that's why we get $$\frac{d}{dx}x=\frac{d}{dx}f^{-1}(f(x)).$$

In the expression $$\frac{df^{-1}}{dx}(f(x))f'(x)$$ we are multiplying two functions, $x\mapsto\frac{df^{-1}}{dx}(f(x))$ and $x\mapsto f'(x)$.

(Remember, to multiply two functions $x\mapsto g(x)$ and $x\mapsto h(x)$, we send each $x$ to the product of their two values, making the function $x\mapsto g(x)\cdot h(x)$.)

In the expression $$\frac{df^{-1}}{dx}(f(x)),$$ the symbols $\frac{df^{-1}}{dx}$ denote the function that is the derivative of $f^{-1}$, and we are plugging in the value $f(x)$ to this function. Thus, $$\frac{df^{-1}}{dx}(f(x))$$ is the derivative of $f^{-1}$ at the point $f(x)$ (just like the derivative of $x^2$ at the point $3$ is $6$).

The symbol $f'(x)$ is of course another way of writing the derivative of $f$ at the point $x$, i.e. $$f'(x)=\frac{df}{dx}(x).$$

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Thanks. That was a very good answer. –  user3200098 Feb 22 at 19:11

Simply by the chain rule you have the third equality and then you deduce the expression of the derivative $(f^{-1})'(f(x))$ by division.

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See $f^{-1}(f(x))$ as a function with argument $f(x)$. Than, if you want to find $df^{-1}(f(x))/dx$ you must use the chain rule, i.e. $df^{-1}(f(x))/dx=[df^{-1}(f(x))/df(x)]*[df(x)/dx]$.

Than from $1=df^{-1}(f(x))/dx * f'(x)$ we have that $df^{-1}(f(x))/dx=1/f'(x)$.

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The chain rule is used when you differentiate a function of a function of $x$, when $x$ is the variable with respect to which you're differentiating. That's what you got where you see $\dfrac{d}{dx}f^{-1}(f(x))$.

$$ 1 = \dfrac{d}{dx} x = \frac{d}{dx} f^{-1}(f(x)) = {f^{-1}}^\prime (f(x))\cdot f'(x). $$

Therefore $$ \frac{1}{f^{\,\prime}(x)} = {f^{-1}}^\prime(f(x)). $$

I'd go on from there to write

$$ \frac{1}{f^{\,\prime}(f^{-1}(y))} = {f^{-1}}^\prime (y). $$

That tells you how to differentiate inverse functions and shows you how that comes from the chain rule.

Here's how I would present it. The chain rule can be stated as $$ \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}. $$ That treats $du$ as if it were literally a number. So $$ 1 = \frac{dw}{dw} = \frac{dw}{dx}\cdot\frac{dx}{dw}. $$ Therefore $$ {f^{-1}}^\prime(x) = \frac{dw}{dx} = \frac{1}{dx/dw} = \frac{1}{\frac{d}{dw}f(w)} = \frac{1}{f^\prime(w)} = \frac{1}{f^\prime(f^{-1}(x))}.$$

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