Differentiating Inverse Trigonometric Functions (DP IB Analysis & Approaches (AA)): Revision Note

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Paul

Last updated

5 June 2025

Differentiating Inverse Trigonometric Functions

What are the inverse trigonometric functions?

arcsin, arccos and arctan are functions defined as the inverse functions of sine, cosine and tangent respectively
- $\arcsin (\frac{\sqrt{3}}{2}) = \frac{π}{3}$ which is equivalent to $\sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$
- $\arctan (- 1) = \frac{3 π}{4}$ which is equivalent to $\tan (\frac{3 π}{4}) = - 1$

What are the derivatives of the inverse trigonometric functions?

$f (x) = \arcsin x$
- $f' (x) = \frac{1}{\sqrt{1 - x^{2}}}$
$f (x) = \arccos x$
- $f' (x) = - \frac{1}{\sqrt{1 - x^{2}}}$
$f (x) = \arctan x$
- $f' (x) = \frac{1}{1 + x^{2}}$
Unlike other derivatives these look completely unrelated at first
- their derivation involves use of the identity $\cos^{2} x + \sin^{2} x \equiv 1$
- hence the squares and square roots!
All three are given in the formula booklet
Note with the derivative of $\arctan x$ that $(1 + x^{2})$ is the same as $(x^{2} + 1)$

How do I show or prove the derivatives of the inverse trigonometric functions?

For $y = \arcsin x$
- Rewrite, $\sin y = x$
- Differentiate implicitly, $\cos y \frac{d y}{d x} = 1$
- Rearrange, $\frac{d y}{d x} = \frac{1}{\cos y}$
- Using the identity $\cos^{2} y \equiv 1 - \sin^{2} y$ rewrite, $\frac{d y}{d x} = \frac{1}{\sqrt{1 - \sin^{2} y}}$
- Since, $\sin y = x$ , $\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^{2}}}$
Similarly, for $y = \arccos x$
- $\cos y = x$
- $- \sin y \frac{d y}{d x} = 1$
- $\frac{d y}{d x} = - \frac{1}{\sin y}$
- $\frac{d y}{d x} = - \frac{1}{\sqrt{1 - \cos^{2} y}}$
- $\frac{d y}{d x} = - \frac{1}{\sqrt{1 - x^{2}}}$
Notice how the derivative of $y = \arcsin x$ is positive but is negative for $y = \arccos x$
- This subtle but crucial difference can be seen in their graphs
  - $y = \arcsin x$ has a positive gradient for all values of $x$ in its domain
  - $y = \arccos x$ has a negative gradient for all values of $x$ in its domain

What do the derivative of inverse trig look like with a linear function of x?

For linear functions of the form $a x + b$
$f (x) = \arcsin (a x + b)$
- $f' (x) = \frac{a}{\sqrt{1 - {(a x + b)}^{2}}}$
$f (x) = \arccos (a x + b)$
- $f' (x) = \frac{a}{\sqrt{1 - {(a x + b)}^{2}}}$
$f (x) = \arctan (a x + b)$
- $f' (x) = \frac{a}{1 + {(a x + b)}^{2}}$
These are not in the formula booklet
- they can be derived from chain rule
- they are not essential to remember
- they are not commonly used

Examiner Tips and Tricks

For $f (x) = \arctan x$ the terms on the denominator can be reversed (as they are being added rather than subtracted)
- $f' (x) = \frac{1}{1 + x^{2}} = \frac{1}{x^{2} + 1}$
- Don't be fooled by this, it sounds obvious but on awkward "show that" questions it can be off-putting!

Worked Example

a) Show that the derivative of $\arctan x$ is $\frac{1}{1 + x^{2}}$

b) Find the derivative of $\arctan (5 x^{3} - 2 x)$ .

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Previous:Differentiating Reciprocal Trigonometric FunctionsNext:Differentiating Exponential & Logarithmic Functions

Differentiating Inverse Trigonometric Functions (DP IB Analysis & Approaches (AA)): Revision Note

Differentiating Inverse Trigonometric Functions

What are the inverse trigonometric functions?

What are the derivatives of the inverse trigonometric functions?

How do I show or prove the derivatives of the inverse trigonometric functions?

What do the derivative of inverse trig look like with a linear function of x?

You've read 0 of your 5 free revision notes this week

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

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