The Calculus of Inverses

Today we will consider computing the derivative of the inverse of a function. This is pretty standard and is in all the textbooks.

The usual suspects are the inverse trigonometric functions. So let’s start with $y={{\sin }^{-1}}\left( x \right)$ and then rewrite this as $x=\sin \left( y \right)$ . Differentiating this gives

$\displaystyle 1=\cos \left( y \right)\frac{dy}{dx}$
$\displaystyle \frac{dy}{dx}=\frac{1}{\cos \left( y \right)}$

Since we would like this in terms of x we can proceed two ways.

The denominator is the cosine of the number whose sine is x. So using the relationship

$\cos \left( y \right)=\sqrt{1-{{\sin }^{2}}\left( y \right)}=\sqrt{1-{{x}^{2}}}$

we find that

$\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-{{x}^{2}}}}$ .

That tends to be confusing so another method is to draw a right triangle with an acute angle of y and arrange the side so that

$\displaystyle \sin \left( y \right)=\frac{x}{1}=\frac{\text{opposite}}{\text{hypotenuse}}$

From this we can find all the trigonometric functions of y, specifically:

$\displaystyle \cos \left( y \right)=\frac{\text{adjacent}}{\text{hypotenuse}}-\frac{\sqrt{1-{{x}^{2}}}}{1}=\sqrt{1-{{x}^{2}}}$ and $\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-{{x}^{2}}}}$ .

A second example: Find the derivative of $y={{\sec }^{-1}}\left( x \right)$ . The domain of this function is $\left| x \right|\ge 1$ and the range is $[0,\tfrac{\pi }{2})\cup (\tfrac{\pi }{2},\pi ]$ , the function is increasing on both parts of its domain; we will need to know this.

Proceeding as above we will find that

$\displaystyle \frac{dy}{dx}=\frac{1}{\sec \left( y \right)\tan \left( y \right)}$ .

Drawing a triangle as above and arranging the side so that sec(y) = x:

Then $\displaystyle \frac{dy}{dx}=\frac{1}{x\sqrt{1+{{x}^{2}}}}$ ,

But wait! It may be that x < 0, but ${{\sec }^{-1}}\left( x \right)$ is increasing and the derivative should always be positive. So, this needs to be adjusted to

$\displaystyle \frac{d}{dx}{{\sec }^{-1}}\left( x \right)=\frac{1}{\left| x \right|\sqrt{1+{{x}^{2}}}}$

These can be a bit tricky.

Next: the fifth and last posting in this series will look at the graphical and numerical aspects of the derivatives of a function and its inverse.

3 thoughts on “The Calculus of Inverses”

No, it is possible to pick a range for inverse secant so that it is *not* increasing on both pieces, precisely so that the derivative formula needs no absolute value (and indeed cannot have one). I believe that the usual choice under these circumstances is [0, pi/2) union [pi, 3pi/2) for the inverse secant.

I don’t advocate this, but it does have its advantage: no absolute value.

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You are correct. Over the years the ranges of the inverse trig functions have been given differently in different books. WolframAlpha gives [0, pi/2) union (pi/2, pi], which I think is what most books are using now. Regardless of what range you choose the function is increasing on both halves, therefore the derivative needs the absolute value. Similar considerations are needed for the other inverse trig functions especially arccsc(x) which is decreasing on both parts of its domain [-pi/2, 0) union (0, pi/2], so it’s derivative is negative everywhere.

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Lin

It seems that I have seen conflicting statements about the domain for the inverse secant formula. Am I mistaken or is there some lack of consistency in this issue?

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Teaching Calculus

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The Calculus of Inverses

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