Differentiation of Inverse Functions

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[edit] Motivation

Knowing this theorem is useful in many ways. It is, in fact, a source of obscure tricks, but it also gives a succinct and effective method for remembering strange derivatives like those of inverse trigonometric functions, the kinds which we encounter so incredibly often in physics. Remembering those derivatives is now reduced to remembering a few core trigonometric identities, and the derivatives of the basic trigonometric functions.

[edit] Theorem

Let $f (x)$ be a function which is differentiable on the open interval $I$ .

If $f' \neq 0$ $(f' \equiv \frac{df(x)}{dx})$ for all $x$ in $I$ then $f$ has an inverse $f - 1$ which is defined and is differentiable on $f (I)$ . That is, for each $a$ in $I$ , $f - 1$ is differentiable at $b = f (a)$ and

$\frac{d}{db} f^{-1}(b) = \left[{\frac{d f(a)}{da}}\right]^{-1}$

[edit] Proof

First, we remark that $f - 1$ is defined, as $f$ is bijective over $I$ . Next, we remark that since $f$ is differentiable, so is $f - 1$ which may be shown simply by swapping the axes of the $f (x)$ graph.

First, parameterizing $a = a (t), b = b (t)$ .

Now, starting from the definition of $b, b = f (a)$ , we must have $a = f - 1 (b)$ . Then

$\frac{da}{dt} = \frac{d}{dt} \left[{f^{-1}(b)}\right] = \frac{d}{db} \left[{f^{-1}(b)}\right] \frac{db}{dt}$

Recalling that $b = f (a)$ ,

$\frac{d}{db} \left[{f^{-1}(b)}\right] \frac{db}{dt} = \frac{d}{db} \left[{f^{-1}(b)}\right] \frac{d f(a)}{dt} = \frac{d}{db} \left[{f^{-1}(b)}\right] \frac{d f(a)}{da} \frac{da}{dt}$

$\frac{da}{dt} = \frac{d}{db} \left[{f^{-1}(b)}\right] \frac{d f(a)}{da} \frac{da}{dt}$

Cancelling $\frac{da}{dt}$ from both sides, and dividing by $\frac{d f(a)}{da}$ which by assumption is non-zero, we arrive at our desired result:

$\frac{d}{db} f^{-1}(b) = \left[{\frac{d f(a)}{da}}\right]^{-1}$

[edit] Uses

As alluded to above, this takes some of the mystery out of those inverse trigonometric derivatives. For example,

Let $y = sinφ$ . Then

$\frac{d \arcsin{y}}{d y} = \left[{\frac{d \sin{\phi}}{d\phi}}\right]^{-1}$ $\frac{d \arcsin{y}}{d y} = \left[{\cos{\phi}}\right]^{-1}$

Referring to our highschool triangle diagram, we see that

$\cos{\phi} = x = \sqrt{1 - y^2}$

Hence, as our first-year calculus teachers asserted,

$\frac{d \arcsin{y}}{d y} = \frac{1}{\sqrt{1 - y^2}}$

Remarkably, the other formulas are similarily derivable. Think of the wasted time you spent fumbling around in the dark for your integral tables! Fumble no more!

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Category: Calculus and Differential Equations