# Derivative of Inverse Functions (a.k.a, How to Create Your Own Table of Derivatives)

Quick question. Are you currently (or have been) a student of **differential calculus** (a.k.a., Calculus I)? If so, maybe you can guess what is on that (potentially-cryptic*)* graph above?

What. Inverse functions of course! Just look at that beautiful shape and how well they correspond with each other!

(OK. Bonus point: can you guess *which* functions they are?)

Yep. There is definitely more than meet the eyes. Granted, **inverse functions** are studied even before a typical calculus course, but their *roles* and *utilities* in the **development of calculus** only start to become increasingly apparent, after the discovery of a certain formula — which related the derivative of an inverse function to its original function. And guess what? *That* — is the topic that we’ll be delving into in greater details today.

So… hang on tight there, as you are about to travel to the land of *exotically-strange functions* in a few seconds. π

# A Primer on Inverse Functions

In theory, given a function $f$ defined on an interval $I$, the role of $f$ is to map $x$ to $f(x)$. That is, $x$ plays the role of the **referrer**, with $f(x)$ being the **target** of $x$. Collectively, the set of all targets of $f$ under $I$ forms the set commonly known as $f(I)$ — or the **image** of $f$ under $I$ (think of the **candle-and-wall analogy** in optics). Under this terminology, the function $f$ is said to map $I$ to $f(I)$.

Presumably, if another function can be constructed to map *each* target back to its referrer, then this new function would be considered the **inverse function** of $f$ — or $f^{-1}$ for short.

Of course. Not all functions can have inverse. However, for a certain class of functions that are deemed **injective** (i.e., functions where every referrer points to a different target), every target will have a *unique* referrer. This means that it would then be possible to define a function — which maps each target in $f(I)$ back to its original referrer in $I$. Such a function would then be rightfully considered the inverse of $f$.

In the context of functions involving real numbers (i.e.,** real-valued functions**), the domains are usually a union of **open** or **close interval**, and in the context of calculus, the functions are generally assumed to be **differentiable** (and hence **continuous**). As it happens, when these assumptions are combined together, the results are a series of *fundamental* and *increasingly powerful* theorems about **invertible functions**:

Given a function $f$ defined on an interval $I$ (possibly with a larger domain), if the function is *injective*Β on $I$, then $f$ — when the domain is restricted to $I$ — has an inverse $f^{-1}$ with domain $f(I)$ .

Given a function $f$ defined on an interval $I$ (possibly with a larger domain), if the function is *continuous* and *injective *on $I$, then $f$ is either **strictly increasing**, or **strictly decreasing** on $I$.

Given a function $f$ defined on an interval $I$ (possibly with a larger domain), if $f$ is *continuous* and *injective* on $I$, then $f^{-1}$ — the inverse of $f$ as defined on $f(I)$ — is continuous throughout $f(I)$ as well. Moreover, if $f$ is strictly increasing (decreasing) on $I$,Β then $f^{-1}$ is strictly increasing (resp., decreasing) on $f(I)$ as well.

A classic example illustrating these theorems would be the function $f(x)=x^2$. For example, when we take the domain of $f$ to be $\mathbb{R}$ (i.e., the set of all **real numbers**), the function $f$ would map both $-1$ and $+1$ to the same number, and is therefore *not* invertible. However, when the domain is *restricted* to the set of **non-negative numbers**, $f$ would now become *injective* everywhere, and hence is by extension *invertible*, with the inverse being the function $f^{-1}(x)=\sqrt{x}$.

Also, since $f$ is *continuous* and *injective* on the set of non-negative numbers, by Theorem 2 $f$ should be **monotone** (or as in this case — strictly increasing) on this domain, and since continuous injective functions produce continuous inverses, the inverse of $f$ — the square root function $\sqrt{x}$ — should be continuous (and strictly increasing) on its own domain as well — which it is!

Note that in general, since the graph of $f$ is the set of points $(x,f(x))$ and $f^{-1}$ the set of points $(f(x),x)$, the two functions are — graphically speaking — **mirror reflection** of each other along the diagonal axis $f(x)=x$.

Alternatively, since $f$ is also the inverse of $f^{-1}$, one can also think of the **graph** of $f^{-1}$ as the set of points $(x, f^{-1}(x))$, and the graph of $f$ the set of points $(f^{-1}(x),x)$. Hence, it is fair to say — in our language at least — that $x$ and $f(x)$ are **correlates** of each other (from the perspective of $f$), much like the same way that $x$ and $f^{-1}(x)$ are correlates of each other (from the perspective of $f^{-1}$).

# Derivative of Inverse Functions

As it stands, mathematicians have long noticed the relationship between a point in a function and *its correlate in the inverse function*. More specifically, it turns out that the **slopes** of tangent lines at these two points are exactly reciprocal of each other! To be sure, here’s a neat animation from Dr. Phan to prove our sanity:

In **real analysis** (i.e., theory of calculus), this geometrical intuition would then constitute the backbone of what came to be known as the **inverse function theorem** (on the derivative of a inverse function), which can be proved using the three aforementioned theorems above:

Given a function $f$ *injectively* defined on an interval $I$ (and hence $f^{-1}$ defined on $f(I)$), $f^{-1}$ is **differentiable** at $x$ if the expression $\frac{1}{f'(f^{-1}(x))}$ makes sense. That is, if the original function $f$ is differentiable at the **correlate **of $x$, with a derivative that is* not *equal to $0$.

In which case, the derivative of $f^{-1}$ at $x$ exists and is equal to the said expression:

\begin{align*} [f^{-1}]'(x) = \frac{1}{f'(f^{-1}(x))} = \left[ \frac{1}{ f'(\Box)}\right]_{\Box = f^{-1}(x)} \end{align*}

In English, this reads:

###### The derivative of an inverse function at a point, is equal to the **reciprocal** of the derivative of the original function — at its correlate.

Math Vault**Leibniz’s notation**:

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

which, although not useful in terms of calculation, embodies the essence of the proof.

# Applications of Inverse Function Theorem

All right. So how do we apply this theorem in practice? Well, for that purpose, here we lay out *7 *examples — for your own pleasure. π

## Example 1 — Linear Functions

If we let the original function to be $f(x)=7x-5$, for instance, then, after solving for $x$ and interchanging the $x$ with $f(x)$, we get that $f^{-1}(x) =\frac{x+5}{7}$. In which case, it is clear that $f'(x)=7$ and $[f^{-1}]'(x)=\frac{1}{7}$.

So, this means that we just found out that the derivatives are reciprocal of each other — without appealing to any higher mathematical machinery.

But let’s say that we were to find $[f^{-1}]'(2)$ using **Inverse Function Theorem**, then here is what we would have to do:

- Find the correlate of $2$
- Calculate the derivative of the original function — at this correlate
- Take the reciprocal

And once that’s done, the number obtained would then be the derivative of the inverse function — at $x=2$.

OK. Let’s see what happens if we follow through the steps:

- $f^{-1}(2) =\left[ \frac{x+5}{7} \right]_{x=2} = 1$, so the correlate of $2$ is $1$ (i.e., $2$ is to $f^{-1}$ as $1$ is to $f$).
- $f'(1)=\left[ 7\right]_{x=1} = 7$
- Taking the reciprocal, we get $\frac{1}{7}$, so that $[f^{-1}]'(2) =\frac{1}{7}$ — as expected from our result before..

Of course. In this case, **Inverse Function Theorem** is not really necessary, but it does illustrate the mechanics of calculating the derivative of an inverse function fairly well to get the momentum going.

## Example 2 — Square Root Function

Now that we know that **square** and **square root** functions are inverses of each other (when we restrict the domain of $f(x)=x^2$Β to the *non-negative numbers*), we should be able to calculate the derivative of the square root function with the help of its inverse. To illustrate, here’s how we can find the derivative of $f^{-1}(x)=\sqrt{x}$ at $5$:

- Correlate of $5$? Just $f^{-1}(5)=\sqrt{5}$. So $5$ is to $f^{-1}(x)=\sqrt{x}$ as $\sqrt{5}$ is to $f(x)=x^2$.
- Derivative of the original function at $\sqrt{5}$? $f'(\sqrt{5})=\left[ 2x\right]_{x=\sqrt{5}}=2\sqrt{5}$.
- Reciprocal? $\frac{1}{2\sqrt{5}}$.

Therefore, $(\sqrt{x})'(5)=\frac{1}{2\sqrt{5}}$ — exactly as one would expect from the use of **power rule**.

So all is good in this case. However, don’t commit the *non-mathematician-like* mistake of indiscriminately invoking **Inverse Function Theorem**Β when the preconditions are not satisfied.

In particular, the derivative of $\sqrt{x}$ does not exist at $x=0$, because derivative of the original function at its correlate is $0$, so that if you choose to take the reciprocal anyway, your derivative will blow up big time. Why? Because the square root function actually has a **vertical tangent** at $x=0$!

However, when $x>0$, $(\sqrt{x})’=\frac{1}{\left[ 2x\right]_{x=\sqrt{x}}}=\frac{1}{2\sqrt{x}}$, which illustrates that while the derivative of $x^2$ can be proved from definition, the derivative of $\sqrt{x}$ can be proved — with a bit more style — using **Inverse Function Theorem**!

## Example 3 — General Root Functions

While $x^n$ and $\sqrt[n]{x}$ are inverses of each other alright ($n \in \mathbb{N}, n\ge2$), it still needs to be recognized that there are really *two kinds* of root functions: one where the root is an **even** (natural) number, and one where the root is **odd**.

Indeed, when the root is *even*, $x^n$ and $\sqrt[n]{x}$ are inverses of each other *only* insofar as the domain of $f(x)=x^n$ is restricted to **non-negative numbers**, but when the root is *odd*, these two functions are inverses of each other under the full domain of $f$ (i.e., $\mathbb{R}$).

Using the same steps as before, we can find the derivatives of $f^{-1}(x)=\sqrt[n]{x}$ at $x$ for the appropriate domains:

- Correlate of $x$: just $f^{-1}(x)$, or $\sqrt[n]{x}$.
- Derivative of the original function at the correlate: $\left[ nx^{n-1}\right]_{x=\sqrt[n]{x}}=n (\sqrt[n]{x})^{n-1}$
- Reciprocal: $\frac{1}{n (\sqrt[n]{x})^{n-1}} = \frac{1}{n}\frac{1}{x^{(1-\frac{1}{n})}} = \frac{1}{n}x^{(\frac{1}{n}-1)}$

So for the **even** root functions, $ (\sqrt[n]{x})’=\frac{1}{n}x^{(\frac{1}{n}-1)}$ for $x>0$, and for the **odd** root functions, the same is true for $x\ne0$ (why?). Either way, the derivative is left undefined at $x=0$.

For example, $(\sqrt[10]{x})’= \frac{1}{10}x^{\frac{1}{10}-1}$ for $x>0$, and $(\sqrt[3]{x})’=\frac{1}{3}x^{\frac{1}{3}-1}$ for all $x \ne 0$.

And as with before, the formula for the derivative of root functions is really just an special instance of the **power rule**. In fact, one way to think about it, is that this is how the **power rule for derivatives** came about — the derivative of $x^n$ proved from definitions, and the derivative of $\sqrt[n]{x}$ proved using **Inverse Function Theorem**.

## Example 4 — Logarithmic functions

Depending on which statements are adopted as definitions, the derivative of **logarithmic functions** either follows immediately — as a corollary from the definition of logarithm, or is proved using the derivative of its inverse — the **exponential function**.

Under the natural base $e$, the exponential function $f(x)=e^x$ on $\mathbb{R}$ implicitly* defines* the logarithmic function $f^{-1}(x)=\ln{x}$ on $\mathbb{R}_+$ (i.e., the set of **positive numbers**). Using **Inverse Function Theorem**, the derivative of $\ln{x}$ at any $x>0$ can be calculated as follows:

- Correlate of $x$: $f^{-1}(x)=\ln{x}$
- Derivative of the original function at the correlate: $\left[ e^x\right]_{x=\ln{x}} = e^{\ln{x}} = x$
- Reciprocal: $\frac{1}{x}$

Voila! $(\ln{x})’= \frac{1}{x}$ for $x>0$, as expected.

Wait…what about logarithmic functions of any *arbitrary* base $b$ ($b>0$, $b \ne 1$)? Well, the **change of base theorem** for logarithms tells us that:

$$\log_b(x)= \frac{\ln{x}}{\ln{b}}$$

Therefore, for $x>0$:

$$ [\log_b(x)]’ = \left[\frac{\ln{x}}{\ln{b}}\right]’ = \frac{1}{\ln{b}}(\ln{x})’= \frac{1}{x\ln{b}} $$

For example, $[\log_2(x)]’= \left[\frac{\ln{x}}{\ln{2}}\right]’=\frac{1}{x\ln{2}}$, and $[\log_{10}(x)]’=\frac{1}{x\ln{10}}$.

So all is still good. Moving on!

## Example 5 — Inverse Trigonometric Function: Arcsine

For an angle $\theta$, $\sin{\theta}$ denotes the **y-coordinate** of a terminal point with angle $\theta$. For example, $\sin{(-\frac{\pi}{2})}=-1$, $\sin{\frac{\pi}{2}}=1$, and when $\theta$ is between $\frac{-\pi}{2}$ and $\frac{\pi}{2}$, $\sin{\theta}$ is also squeezed between $-1$ and $+1$.

In fact, it’s not too hard to visualize that as $\theta$ moves from $\frac{-\pi}{2}$ to $\frac{\pi}{2}$, $\sin{\theta}$ increases strictly non-stop from $-1$ to $+1$. This means that the function $f(\theta)=\sin{\theta}$ is *injective* on $I = [\frac{-\pi}{2}, \frac{\pi}{2}]$, and by extension, an inverse function of sine — denoted as $\arcsin(x)$ or $\sin^{-1}(x)$ — can be defined on $f(I) = [-1, 1]$.

Naturally, this leads to a series of *calculus-related* questions:

- Where is this inverse sine function differentiable?
- What is the derivative in that case?

Luckily, this is where **Inverse Function Theorem** comes to the rescue big time:

- Correlate of $x$: simply $\sin^{-1}(x)$
- Derivative of the original function at the correlate: $\left[ \cos{x} \right]_{x=\sin^{-1}(x)}= \cos{(\sin^{-1}(x))}$

OK. Before we move on, let’s simplify the expression a little bit. By **Pythagorean Theorem**, we know that $\cos^2{\theta}=1 – \sin^2{\theta}$, or that $\cos \theta = \pm \sqrt{1 – \sin^2 \theta}$. In particular:

\[ \cos{\theta}=\sqrt{1 – \sin^2{\theta}} \qquad (\text{if } \cos{\theta}\ge 0) \]

(e.g., if the argument of cosine ranges from $\frac{-\pi}{2}$ to $\frac{\pi}{2}$ (i.e., stays on the *right* quadrants))

Fortunately, since $\frac{-\pi}{2} \le \sin^{-1}(x) \le \frac{\pi}{2}$ for all $x$ by construction, the aforementioned expression becomes:

$$ \cos{[\sin^{-1}(x)]} = \sqrt{1 – \sin^2[\sin^{-1}(x)]} = \sqrt{1-x^2}$$

### Careful!###### OK. That was pretty clean isn’t it? But that last equality need not hold if the domain of $\sin{x}$ was restricted differently. For example, if the domain of $\sin{x}$ were to be restricted to $[\frac{\pi}{2},\frac{3\pi}{2}]$, $\sin^{-1}(x)$ would have mapped $[-1,1]$ to $[\frac{\pi}{2},\frac{3\pi}{2}]$, which means that $ \cos{[\sin^{-1}(x)]} = \sqrt{1 – \sin^2(\sin^{-1}(x))}$ would no longer apply. For this reason, $[\frac{-\pi}{2},\frac{\pi}{2}]$ is usually regarded as the **standard restricted domain** for sine — when it comes to defining the inverse of sine.

###### OK. That was pretty clean isn’t it? But that last equality need not hold if the domain of $\sin{x}$ was restricted differently. For example, if the domain of $\sin{x}$ were to be restricted to $[\frac{\pi}{2},\frac{3\pi}{2}]$, $\sin^{-1}(x)$ would have mapped $[-1,1]$ to $[\frac{\pi}{2},\frac{3\pi}{2}]$, which means that $ \cos{[\sin^{-1}(x)]} = \sqrt{1 – \sin^2(\sin^{-1}(x))}$ would no longer apply. For this reason, $[\frac{-\pi}{2},\frac{\pi}{2}]$ is usually regarded as the **standard restricted domain** for sine — when it comes to defining the inverse of sine.

All right. That finishes Step 2, which means that when $x=-1$ or $x=-1$, $\sin^{-1}(x)$ is not differentiable as $\sqrt{1-x^2}$ would be $0$. On the other side of the token, it also means that $\sin^{-1}(x)$ *would* be differentiable anywhere on the interval $(-1,1)$. Let’s finish up the third step in calculating the derivative of inverse sine then:

3. Reciprocal: $\frac{1}{\sqrt{1-x^2}}$

Now, putting everywhere together, we have that $[\sin^{-1}]'(x) = \frac{1}{\sqrt{1-x^2}}$ for $-1<x<1$. This means that while $\sin^{-1}(x)$ is defined and *continuous* on $[-1,1]$, it is only differentiable on the interval $(-1,1)$.

To be sure, here’s an accompanying graph of $\sin{x}$ and $\arcsin{x}$ — for the record:

## Example 6 — Inverse Trigonometric Function: Arccosine

For an angle $\theta$, $\cos{\theta}$ denotes the **x-coordinate** of a terminal point with angle $\theta$. For example, $\cos{0}=1$, $\cos{\pi}=-1$, and when $\theta$ is between $0$ and $\pi $, $\cos{\theta}$ is also squeezed between $-1$ and $+1$.

Similar to the case with the **sine** function, it’s not too hard to visualize that as $\theta$ moves from $0$ to $\pi$, $\cos{\theta}$ decreases strictly from $+1$ to $-1$ non-stop. This means that the function $f(\theta) = \cos{\theta}$ is *injective* on $I = [0,\pi]$, and by extension, an inverse function of cosine — denoted by $\arccos(x)$ or $\cos^{-1}(x)$ — can be defined on $f(I) = [-1, 1]$.

So, as constructed above, when is $\cos^{-1}(x)$ differentiable? And what would the derivative be in that case? Well, to find out more, let’s invoke **Inverse Function Theorem** again:

- Correlate of $x$: just $\cos^{-1}(x)$
- Derivative of the original function at the correlate: $\left[ -\sin{x} \right]_{x=\cos^{-1}(x)} = -\sin{(\cos^{-1}(x))}$

And here’s the **fancy Pythagorean trick** again! First, $\sin^2{\theta}=1 – \cos^2{\theta}$, so that $\sin \theta = \pm\sqrt{1 – \cos^2{\theta}}$. In particular:

\[ \sin{\theta}=\sqrt{1 – \cos^2{\theta}} \text{ , if } \sin{\theta}\ge 0 \]

(e.g., if the argument of sine ranges from $0$ to $\pi$ (i.e., remains on the *upper* quadrants))

Luckily, since by construction $\cos^{-1}(x)$ ranges from $0$ to $\pi$, the expression from Step 2 can be simplified as follows:

$$ – \sin{[\cos^{-1}(x)]} = – \sqrt{1 – \cos^2[\cos^{-1}(x)]} = – \sqrt{1-x^2}$$

By inspection, this means that when $x$ is equal to $+1$ or $-1$, $\cos^{-1}(x)$ is not differentiable as $- \sqrt{1-x^2}$ would be $0$. However, the good news is that $\cos^{-1}(x)$ *would* be differentiable anywhere on $(-1,1)$. Finishing up Step 3, we get that:

3. Reciprocal: $\dfrac{1}{- \sqrt{1-x^2}}$

which shows that $[\cos^{-1}]'(x) = \dfrac{1}{-\sqrt{1-x^2}}$ for $-1<x<1$.

In other words, while arccosine is defined and *continuous* on $[-1,1]$, it is only differentiable on $(-1,1)$. Similiar to the arcsine function, arccosine also has vertical tangents at $x=-1$ and $x=1$.

And before we move on, let’s just mention that in general, the **standard restricted domain** for cosine is $[0,\pi]$ when it comes to defining inverse cosine, so that by convention, $\cos^{-1}(x)$ is typically understood to be *the* inverse cosine function mapping $[-1,1]$ to $[0,\pi]$. Were $\cos^{-1}(x)$ to be defined even slightly differently, the **fancy Pythagorean trick** introduced earlier could either *cease to work out properly* — or lead to a new, **non-standard formula** for the derivative of inverse cosine.

## Example 7 — Inverse Trigonometric Functions: Arctangent

GIven an angle $\theta$, $\tan{\theta}$ can be defined as the **ratio** between the y and x-coordinates of the terminal point with angle $\theta$. Geometrically, $\tan{\theta}$ in essence measures the **slope** between the origin, and the terminal point with angle $\theta$.

For example, $\tan{0}=0$, because when the angle is $0$, so does the slope. On the other hand, $\tan{\frac{\pi}{4}}=1$, because when the angle is $45^{\circ}$ degrees, the slope is $1$.

In general, we interpret the **standard restricted domain** of the tangent function to be the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. Under that interpretation, one can see that the *bigger* the angle, the *more positive* the slope. More specifically, as the angle moves from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, the tangent (i.e., slope) goes from *negatively sloppy*, *flat*, to *positively sloppy beyond bound*.

Algebraically, this means that the function $f(\theta) = \tan{\theta}$ is strictly increasing on $(-\frac{\pi}{2}, \frac{\pi}{2})$ as it travels from $-\infty$ to $+\infty$, making it *injective* on $I = (-\frac{\pi}{2}, \frac{\pi}{2})$. As a result, a inverse function of $\tan{\theta}$ — usually denoted by $\arctan{x}$ or $\tan^{-1}(x)$ — can be defined on $f(I) = (-\infty,\infty)$.

Hence by construction, $\tan^{-1}(x)$ has this *neat* property that it is defined everywhere and *continuous* on $\mathbb{R}$. Interesting! So how does $\tan^{-1}(x)$ behaves in terms of differentiability and derivatives then?

Well, there is one way we can find that out — through **Inverse Function Theorem** of course!

- Correlate of $x$: $\tan^{-1}(x)$
- Derivative of the original function at the correlate: $\left[ \sec^2(x)\right]_{x=\tan^{-1}(x)}=\sec^2(\tan^{-1}(x))$

Now, to prevent us from *blindly manipulating symbols*, let’s just make sure that things still make sense here. We are talking about the derivative of $\tan^{-1}(x)$, so $x$ could stand for any real number from $-\infty$ to $+\infty$. By construction, $-\frac{\pi}{2} < \tan^{-1}(x) < \frac{\pi}{2}$Β (i.e., stays on the right side of the **right quadrants**). As a result, $\sec^2(\tan^{-1}(x))$ is hence well-defined everywhere on $\mathbb{R}$ — regardless of the value of $x$. Oh well, guess this means that everything is still on track!

Also, since we know that $\sec^2{\theta} = \tan^2{\theta} + 1$ for $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, the aforementioned expression becomes:

\[ \sec^2(\tan^{-1}(x)) = 1 + \tan^2(\tan^{-1}(x)) = 1 + x^2 \]

which is really good, because $1+x^2 \ge 1 \ne 0$ for all $x \in \mathbb{R}$. This means that the function $\tan^{-1}(x)$ is differentiable *everywhere* in its domain. Finishing Step 3 quickly yields:

Β Β Β Β Β Β Β 3. Reciprocal: $\dfrac{1}{1+x^2}$

which means that $[\tan^{-1}]'(x) = \dfrac{1}{1+x^2}$ for any $x \in \mathbb{R}$. Perfection!

And to reward ourselves with some “math toys” for the great work being done, here’s a graph of both $\tan{x}$ and $\tan^{-1}(x)$ — as defined using $\tan{x}$’s **standard restricted domain**:

# Closing Words and Beyond

So…is there more to this **Inverse Function Theorem** thingy? You bet! In particular, here are some recommended functions you might be tempted to differentiate — and add to your own list of favorite derivatives:

- $\sec^{-1}(x)$
- $\csc^{-1}(x)$
- $\cot^{-1}(x)$
- $\sinh^{-1}(x)$ (note: $\sinh{x}= \dfrac{e^x – e^{-x}}{2}$)
- $\ldots$

(heck, why not just construct your own **injectively differentiable** function from scratch, and find the derivative of its inverse?)

Either way, the take-home lesson — if there is any — is that by acknowledging that some functions come in pairs, and by learning how to *maneuver* around the formula for the derivative of an inverse, one can double the number of functions in their own **table of derivatives**, thereby doubling their knowledge on derivatives in general.

And before we go, here is a summary of what we have *invented*/*discovered* so far:

For all $x \in \mathbb{R}$, $[\tan^{-1}]'(x)=\frac{1}{1+x^2}$.

OK. Enough of inverse functions for now. Tune in next time for more math goodies perhaps? In the meantime, you can drop by our Facebook and satisfy your craving for more **edutaining tidbits** and “**math cookies**“. π

Derek Pareto

March 3, 2016 @ 3:39 pm

Oh man. I wish they presented it the way you did when I learnt Cal 1. Excellent work!

Math Vault

March 3, 2016 @ 7:41 pm

Glad you enjoy it. Thanks!

Anitej Banerjee

May 18, 2016 @ 10:05 pm

This is amazing!

This site has been so helpful in making Calculus easier for me π (I hate having to remember the derivatives of arcsec and such functions, and I’m loving all the e and ln(x) derivations you guys do!)

Kudos to the writers! π

Math Vault

May 19, 2016 @ 12:30 am

Thank you! More derivative goodies coming up this month! π