# Derivative of inverse function using $dx/dy$ - what is going on here?

I'm about to learn about differentiating inverse functions at school, and the formula we're being told we'll be using is [assuming $$g(x)=f^{-1}(x)$$]:

$$f'(x)=\frac{1}{g'(f(x))}$$

In other places online, however, I have seen a much simpler formula, $$dx/dy * dy/dx = 1$$. (I am quite interested in Leibniz notation, as the old intuitive ideas often stun me in their elegance.)

However, here I cannot for the life of me figure out the connection between the Leibniz notation above and the more complicated "prime" notation above. I have searched far and wide across the Internet, but am still confused. Can anyone explain? How would I calculate the derivative of an inverse function using Leibniz notation, and how does that connect to the process of computing an inverse function's derivative using "prime" notation?

• If we translate this into $\frac{dx}{dy}$ notation it is like this: $\frac{df}{dx} = \frac{1}{\frac{dg}{dx}(f(x))}$ where the $\frac{dg}{dx}(fx)$ means the derivative of g wrt x evaluated at the point $f(x)$ Commented Nov 25, 2019 at 3:56

Note that $$y=f(x)$$, $$x=g(y)$$ \begin{align} \frac{dy}{dx}\cdot\frac{dx}{dy}&=1\\ \frac{dy}{dx}&=\frac{1}{\frac{dx}{dy}} \\ \frac{df(x)}{dx}=\frac{1}{\frac{dg(y)}{dy}} \\ f'(x)=\frac{1}{g'(f(x))} \end{align}

I think that your teacher needs to clarify the notation with examples.

Definition: Let $$g(x)$$ be injective. Then $$f(x)$$ is an inverse function of $$g(x)$$ provided

$$g(f(x)) = f(g(x)) = x$$

Example: Suppose $$g(x)=2x-1$$, which is injective. Then, $$g^{-1}(x)$$ can be found by switching $$x$$ and $$y$$ and then solving for $$y$$

$$x=2y-1\implies y=\frac{x+1}{2} \implies g^{-1}(x)=\frac{x+1}{2}$$

Alternatively, if we define $$f(x)=\dfrac{x+1}{2}$$ then

$$g(f(x))=g\left(\frac{x+1}{2}\right)=2\left(\frac{x+1}{2}\right)-1=x$$

and

$$f(g(x))=f\left(\frac{(2x-1)+1}{2}\right)=x$$

so we see that the definition holds in this example. This leads to the result provided by your course notes.

Theorem: Let $$g(x)$$ be a function that is differentiable on an interval $$I$$. If $$g(x)$$ has an inverse function $$f(x)$$, then $$f(x)$$ is differentiable at any $$x$$ for which $$g'(f(x))\neq 0$$. Moreover,

$$f'(x)=\frac{1}{g'(f(x))},~~g'(f(x))\neq 0$$

$$g(f(x)) = x$$

and then differentiate implicitly

$$\frac{d}{dx}\Big[g(f(x))\Big]=\frac{d}{dx}(x)$$

where we set $$y=g(u)$$ and $$u=f(x)$$ and then use the chain rule

$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=g'(u)f'(x)=g'(f(x))f'(x)$$

and since $$\frac{d}{dx}(x)=1$$, we know that

$$g'(f(x))f'(x)=1$$

then since $$g'(f(x))\neq 0$$, we can divide by it to form

$$f'(x)=\frac{1}{g'(f(x))}$$

Example: Suppose $$x>0$$ and

$$g(x)=x^2$$

then $$f(x)=\sqrt{x}$$ is its inverse. We have

$$f'(x)=\frac{1}{2\sqrt{x}}$$

and

$$g'(x)=2x$$

therefore

$$g'(f(x))=2(f(x))=2\sqrt{x} \implies \frac{1}{g'(f(x))}=\frac{1}{2\sqrt{x}}$$

so we see that

$$f'(x)=\frac{1}{g'(f(x))}$$

In terms of Leibniz notation, we can adjust the proof so that all of the primes are replaced by differentials. In this notation $$f'(x)=\frac{df(x)}{dx}=\frac{df}{dx}$$

and letting $$y=f(x)$$ and $$x=g(y)$$ forms

$$\frac{dy}{dx}=\frac{df}{dx},~~g'(f(x))=\frac{dg(y)}{dy}=\frac{dx}{dy}$$

so that

$$\frac{dy}{dx}\frac{dx}{dy}=1 \implies \frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$$

or

$$\frac{df}{dx}=\frac{1}{\frac{dg(y)}{dy}}$$

and we once again arrive at

$$f'(x)=\frac{1}{g'(f(x))}$$

as desired.

Example: Suppose that $$y=f(x)=e^x$$. Then, we want to show that

$$f'(x)=\frac{1}{g'(f(x))}$$

We are given

$$\frac{df}{dx}=\frac{dy}{dx}=e^x$$

Next, we can deduce that $$x=g(y)=\ln(y)$$ so that $$g$$ is the natural log function (the inverse of $$e^x$$ is the natural log function). Then

$$g'(f(x))=\frac{dx}{dy}=\frac{dg(y)}{dy}=\frac{d}{dy}\ln(y)=\frac{1}{y}=\frac{1}{e^x}$$

Therefore,

$$\frac{1}{g'(f(x))}=e^x$$

In terms of prime notation versus Leibniz notation, it is largely a matter of personal preference. Many people prefer using the Leibniz notation because the chain rule can be easily identified as

$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$

$$\Big[f(g(x))\Big]' = f'(g(x))g'(x)$$

$$\frac{dy}{dx}$$

makes it abundantly clear that $$x$$ is the independent variable and $$y$$ is the dependent variable. This can sometimes be confusing when you have multiple variables with primes.

I often switch between the two notations as needed. It might be best to work with both notations if the primes confuse you.

• This is an excellent answer!! However, could you please add some discussion of $dx/dy$, i.e. how Leibniz notation relates to the prime notation you used? That was also an important part of my question!
– Will
Commented Nov 25, 2019 at 5:00
• Thanks! However, would you be able to add an example (with specific functions) of how you would find the inverse function's derivative using Leibniz notation? I'm still a bit unclear.
– Will
Commented Nov 25, 2019 at 13:37
• It is fine. I need to find time this afternoon to update my answer with examples. Commented Nov 25, 2019 at 16:45
• It is correct that $\dfrac{dy}{dx}\dfrac{dx}{dy}=e^x\left(\dfrac{1}{e^x}\right)=1$. However, the rest of the analysis follows from the fact that $x=g(y)$ so $\dfrac{dx}{dy}=\dfrac{dg(y)}{dy}$. Does the analysis done in the example make sense? Do you have any other examples? Commented Nov 26, 2019 at 2:43
• @Will: That is true in specific scenarios. For example, you can replace $x$ with $y$ when you are integrating a function because the variable of integration is a dummy variable. In this context, the chain rule is the reason why the variable substitution is made. I would go through more examples and review the chain rule. I would also ask your lecturer about any future applications of the chain rule that they make. Commented Nov 26, 2019 at 16:34

$$g(x)=f^{-1}(x)$$ means that $$g(f(x))=x$$. Take the derivative of both sides and we get $$f'(x)g'(f(x))=1$$ by the chain rule. But that's it as $$y=f(x)$$ and $$x=g(y)$$ implies $$dx/dy = g'(y) = g'(f(x))$$ and we're done.

I think this is best explained with physical quantities by interpreting the derivatives in terms of slope of a function. For example the kinetic energy depends on the speed of an object by the formula $$E=\frac{1}{2} m v^2$$. We can write this as $$E = f(v)$$. Suppose the object has mass $$3\ kg$$ and speed $$8\ m s^{-1}$$. Its kinetic energy is $$\frac{1}{2} 3 \times 8^2 = 96\ J$$. Now the slope of $$f$$ at that point is $$24$$ because of the relation $$d E = m v d v$$. It means that an increase of the speed by $$1\$$ metre per second results in a increase of energy of $$24$$ joules. The meaning of the formula is that the slope of the inverse function is the inverse of the slope of $$f$$, that is to say $$\frac{1}{24}$$. Thus an increase of $$1 J$$ of kinetic energy results in an increase of $$\frac{1}{24} m s^{-1}$$ in speed. To sum things up, if $$v=g(E) = f^{-1}(E)$$, we have $$$$f(8)=96\qquad g(96)=8\qquad f'(8) = 24\qquad g'(96) = \frac{1}{24} = \frac{1}{f'(8)} = \frac{1}{f'(g(96))}$$$$ The key point to understand is that we take $$g$$ and $$g'$$ at the point $$96 = f(8)$$, but we take $$f$$ and $$f'$$ at the point $$8 = g(96)$$.