18.01: Single Variable Calculus
Section 5
Implicit differentiation, inverses
Massachusetts Institute of Technology
Last Edit Date: 06/07/2024
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Implicit Differentiation¶
Example 1. $\frac{d}{dx}(x^a) = a x^{a - 1}$.
We provide this by an explicit computation for $a = 0, 1, 2, ...$. From this, we also got the formula for $a = -1, -2, ...$. Let us try to extend this formula to cover rational numbers, as well:
$$a = \frac{m}{n};~~~~~y = x^{\frac{m}{n}}~~~~~\text{where $m$ and $n$ are integers.}$$
We want to compute $\frac{dy}{dx}$. We can say $y^n = x^m$ so $ny^{n - 1}\frac{dy}{dx} = mx^{n - 1}$. Solve for $\frac{dy}{dx}$:
$$\frac{dy}{dx} = \frac{m}{n} \frac{x^{m - 1}}{y^{n - 1}}$$
We know that $y = x^{\left( \frac{m}{n} \right)}$ is a function of $x$.
$$ \begin{aligned} \frac{dy}{dx} &= \frac{m}{n} \left( \frac{x^{m - 1}}{y^{n - 1}} \right) \\ &= \frac{m}{n} \left( \frac{x^{m - 1}}{(x^{m / n})^{n - 1}} \right) \\ &= \frac{m}{n} \frac{x^{m - 1}}{x^{m(n - 1)/n}} \\ &= \frac{m}{n} x^{(m - 1) - \frac{m(n - 1)}{n}} \\ &= \frac{m}{n} x^{\frac{n(m - 1) - m(n - 1)}{n}} \\ &= \frac{m}{n} x^{\frac{nm - n - nm + m}{n}} \\ &= \frac{m}{n} x^{\frac{m}{n} - \frac{n}{n}} \\ \text{So, } \frac{dy}{dx} &= \frac{m}{n} x^{\frac{m}{n} - 1} \end{aligned} $$
This is the same answe as we were hoping to get.
Example 2. Equation of a circle with a radius of 1: $x^2 + y^2 = 1$ which we can write as $y^2 = 1 - x^2$. So $y = \pm \sqrt{1 - x^2}$. Let us look at the positive case:
$$ \begin{aligned} y &= + \sqrt{1 - x^2} = (1 - x^2)^{\frac{1}{2}}\\ \frac{dy}{dx} &= \left( \frac{1}{2} \right) (1 - x^2)^{\frac{-1}{x}} (-2x) = \frac{-x}{\sqrt{1 - x^2}} = \frac{-x}{y} \end{aligned} $$
Now let us do the same thing, using implicit differentiation.
$$ \begin{aligned} x^2 + y^2 &= 1 \\ \frac{d}{dx} \left( x^2 + y^2 \right) &= \frac{d}{dx}(1) = 0 \\ \frac{d}{dx} (x^2) + \frac{d}{dx} (y^2) &= 0 \\ \end{aligned} $$
Applying chain rule in the second term,
$$ \begin{aligned} 2x + 2y \frac{dy}{dx} &= 0 \\ dy \frac{dy}{dx} &= -2x \\ \frac{dy}{dx} &= \frac{-x}{y} \\ \end{aligned} $$
Same answer.
Example 3. $y^3 + xy^2 + 1 = 0$. In this case, it is not easy to solve for $y$ as a function of $x$. Instead, we use implicit differentiation to find $\frac{dy}{dx}$.
$$3y^2 \frac{dy}{dx} + y^2 + 2xy \frac{dy}{dx} = 0$$
We can now solve for $\frac{dy}{dx}$ in terms of y and x.
$$ \begin{aligned} \frac{dy}{dx} (3y^2 + 2xy) &= -y^2 \\ \frac{dy}{dx} &= \frac{-y^2}{3y^2 + 2xy} \\ \end{aligned} $$
Inverse Functions¶
If $y = f(x)$ and $g(y) = x$, we call $g$ the inverse function of $f$, $f^{-1}$:
$$x = g(y) = f^{-1}(y)$$
Now, let us use implicit differentiation to find the derivative of the inverse function.
$$ \begin{aligned} y &= f(x)\\ f^{-1}(y) &= x \\ \frac{d}{dx} (f^{-1}(y)) &= \frac{d}{dx}(x) = 1\\ \end{aligned} $$
By the chain rule:
$$ \begin{aligned} \frac{d}{dy} (f^{-1}(y)) \frac{dy}{dx} &= 1 \\ \text{and} \\ \frac{d}{dy} (f^{-1}(y)) &= \frac{1}{\frac{dy}{dx}}\\ \end{aligned} $$
So, implicit differentiation makes it possible to find the derivative of the inverse function.
Example. $y = \arctan (x)$
$$ \begin{aligned} \tan y &= x \\ \frac{d}{dx} [\tan (y)] &= \frac{dx}{dx} = 1 \\ \frac{d}{dx} [\tan (y)] \frac{dy}{dx} &= 1 \\ \left( \frac{1}{\cos^2 (y)} \right) \frac{dy}{dx} &= 1 \\ \frac{dy}{dx} &= \cos^2 (y) = \cos^2 (\arctan (x)) \\ \end{aligned} $$
This form is messy. Let us use some geometry to simplify it.
In this triangle, $\tan (y) = x$ so
$$\arctan (x) = y$$
The Pythagotian theorem tells us the length of the hypotenuse:
$$h = \sqrt{1 + x^2}$$
From this, we can find
$$\cos (y) = \frac{1}{\sqrt{1 + x^2}}$$
From this, we get
$$\cos^2 (y) = \left( \frac{1}{\sqrt{1 + x^2}} \right)^2 = \frac{1}{1 + x^2}$$
So,
$$\frac{dy}{dx} = \frac{1}{1 + x^2}$$
In other words,
$$\frac{d}{dx} \arctan(x) = \frac{1}{1 + x^2}$$
Graphing an Inverse Function¶
Suppose $y = f(x)$ and $g(y) = f^{-1}(y) = x$. To graph $g$ and $f$ together we need to write $g$ as a function of the variable $x$. If $g(x) = y$, then $x = f(y)$, and what we have done is to trade the variables $x$ and $y$. This is illustrated in the following figure.
| $f^{-1}(f(x)) = x$ | $f^{-1} \circ f(x) = x$ |
| $f(f^{-1}(x)) = x$ | $f \circ f^{-1}(x) = x$ |
