Note that a function and its inverse undo each other. Say $f(x)=2x+1$. The function $f$ takes $x$ and multiplies it by $2$ and then adds $1$. To undo that, we go in reverse orders. We first subtract $1$ from $x$ and then we divide the whole thing by $2$, giving us $f^{-1}(x)=\frac{x-3}{2}$. Here's one more example. Say $f(x)=\sqrt{x+2}$. $f$ takes $x$, adds $2$ and then take the square root. To undo the square root, we square x; and then instead of adding $2$, we subtract $2$. So $f^{-1}(x)=x^2-2$. Last example: $f(x)=\frac{1}{x+4}-2$. $f$ takes $x$, adds $4$, inverses the sum and then minuses $2$. $f^{-1}$ adds 2, inverses the sum and lastly minuses $4$. $f^{-1}(x)=\frac{1}{x+2}-4$
[Here we are using the term "function" loosely. Formally, when one talks about a function, the domain and codomain should be specified. For simplicity, we only provide the expression of the function.]
Differentiating inverse functions
Consider a linear function $l(x)=mx+b$. Its inverse will be $l^{-1}(y)=\frac{1}{m}y-\frac{b}{m}$. We then have $l'(x)=m$ and $(l^{-1})'(y)=\frac{1}{m}$, the reciprocal. We can express this by $\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$.
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| Flipping the graphs preserves tangency. |
Since the line $y=mx+b$ is tangent to the curve on the left and flipping the drawing should preserve this tangency, the line $x=\frac{1}{m}y-\frac{b}{m}$ is the tangent line to $x=f^{-1}(y)$ at $(y_0,x_0)$ and its slope $\frac{1}{m}$ should be the derivative $(f^{-1})'(y_0)$. Since $m=f'(x_0)$, we conclude that $(f^{-1})'(y_0)=\frac{1}{f'(x_0)}=\frac{1}{f'(f^{-1}(y_0))}$, or equivalently $f'(x_0)=\frac{1}{(f^{-1})'(f(x_0))}$.
Examples:
Find the derivative of the inverse of this real function $f(x)=2x+\cos x$.
$f'(x)=2-\sin x$.
$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}=\frac{1}{2-\sin f^{-1}(x)}$.
Let $y=f(x)=x^7+4x-5$. Find $(f^{-1})'(-5)$.
Note that $y=-5$ corresponds to $x=0$. Also, $f'(x)=7x^6+4$, so $(f^{-1})'(-5)=\frac{1}{f'(0)}=\frac{1}{4}$.
Inverse trigonometric functions and their derivatives
Sine and arcsine
The sine function $f: \mathbb{R} \to [-1,1] \; f(x)=\sin x$ is not one-to-one, but if we restrict the domain to be $[-\frac{\pi}{2},\frac{\pi}{2}]$, $f$ becomes one-to-one.
We have $\sin^{-1}y=x \iff \sin x=y \quad \text{and} \quad -\frac{\pi}{2}\leq x \leq \frac{\pi}{2}$.
Then $f^{-1}: [-1,1] \to [-\frac{\pi}{2},\frac{\pi}{2}] \quad f^{-1}(x)=\sin^{-1}x$.
The cancellation equations for inverse functions become
$\sin^{-1}(\sin x)=x \quad -\frac{\pi}{2}\leq x \leq\frac{\pi}{2}$
$\sin(\sin^{-1} x)=x \quad -1 \leq x \leq 1$.
$y=\sin x, -\frac{\pi}{2}\leq x \leq\frac{\pi}{2}$ $y=\sin^{-1} x, -1\leq x \leq 1$
Let $y=\sin^{-1}x$. Then $y'=\frac{1}{\cos y}$ from the formula we discussed earlier. Alternatively, one can differentiate $\sin y=x$ implicitly with respect to x to yield the same result. Now $\cos y \geq 0$ since $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$. We have $\cos y=\sqrt{1-\sin^2{y}}=\sqrt{1-x^2}$, thus $\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}} \quad -1<x<1$.
Cosine and arccosine
The inverse cosine function is handled similarly. The restricted cosine function $f(x)=\cos x, 0\leq x\leq \pi$, is one-to-one and so has an inverse function.
$y=\cos x, 0 \leq x \leq \pi$ $y=\cos^{-1} x, -1\leq x \leq 1$
We have $\cos^{-1}y=x \iff \cos x=y \quad \text{and} \quad 0\leq x \leq \pi$
and $f^{-1}: [-1,1] \to [0,\pi] \quad f^{-1}(x)=\cos^{-1}x$. Its derivative is given by $\frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^2}} \quad -1<x<1$.
Tangent and arctangent
The tangent function can be made one-to-one by restricting its domain to $(-\frac{\pi}{2},\frac{\pi}{2})$.
We have $\tan^{-1}y=x \iff \tan x=y \quad \text{and} \quad -\frac{\pi}{2}\leq x \leq \frac{\pi}{2}$
and $f^{-1}: \mathbb{R} \to (-\frac{\pi}{2},\frac{\pi}{2}) \quad f^{-1}(x)=\tan^{-1}x$.
We know that $\lim\limits_{x \to \frac{\pi}{2}^-} \tan x=\infty$ and $\lim\limits_{x \to -\frac{\pi}{2}^+} \tan x=-\infty$, namely, $x=\pm \frac{\pi}{2}$ are vertical asymptotes of the graph of $\tan x$. The graph of $\tan^{-1}$ is obtained by reflecting the graph of the restricted tangent function about $y=x$, it follows that $y=\pm \frac{\pi}{2}$ are horizontal asymptotes of the graph of $\tan^{-1}$. We thus have $\lim\limits_{x \to \infty} \tan^{-1} x=\frac{\pi}{2}$ and $\lim\limits_{x \to -\infty} \tan^{-1} x=-\frac{\pi}{2}$.
$y=\tan x, -\frac{\pi}{2}\leq x \leq\frac{\pi}{2}$ $y=\tan^{-1} x, x \in \mathbb{R}$
The derivative of the arctangent function is $\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2} \quad -1<x<1$.
Inverse Hyperbolic Function and Their Derivatives
Inverse function integration
Suppose $f$ is a continuous one-to-one function. Then $\int f^{-1}(x)dx=xf^{-1}(x)-\int f(u) d(u)=xf^{-1}(x)-F(f^{-1}(x)).$
Definite integral version
$\int_a^b f^{-1}(x)=[xf^{-1}(x)]_a^b-\int_{f^{-1}(a)}^{f^{-1}(b)} f(u) du=[xf^{-1}(x)]_a^b-F(f^{-1}(b))+F(f^{-1}(a))$
Examples:
$\begin{align} \int \arctan x \:dx &= x\arctan x-\int \tan u \:du \quad [u=\arctan x]\\
&=x\arctan x+\ln|\cos u|+C\\ &=x\arctan x-\frac{1}{2}\ln(1+x^2)+C \end{align}$
since $\ln|\cos u|=\frac{1}{2}\ln(\cos^2 u)=-\frac{1}{2}\ln(\sec^2 u)=-\frac{1}{2}\ln(1+x^2)$.
$\int \ln x \:dx=x\ln x-x+C$
$\int \arccos x \:dx=x\arccos x-\sin(\arccos x)+C$
Here's an interesting theorem.
If $f(a)=c$ and $f(b)=d$, we have $\int_c^d f^{-1}(y)dy\:\text{[blue region]}+\int_a^b f(x)dx\:\text{[grey region]}=bd-ac$.
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| Proof without words |
Related
Reference:
Differentiating inverse functions
Inverse trigonometric functions and their derivatives
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