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

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

  •  $\arcsin \left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{3}$ which is equivalent to $\sin \left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}$

  •  $\arctan (-1)=\frac{3\pi }{4}$ which is equivalent to $\tan \left(\frac{3\pi }{4}\right)=-1$

What are the derivatives of the inverse trigonometric functions?

• $f\left(x\right)=\arcsin x$

  • $f^{\text{'}}\left(x\right)=\frac{1}{\sqrt{1-x^2}}$

• $f\left(x\right)=\arccos x$

  • $f^{\text{'}}\left(x\right)=-\frac{1}{\sqrt{1-x^2}}$

• $f\left(x\right)=\arctan x$

  • $f^{\text{'}}\left(x\right)=\frac{1}{1+x^2}$

• Unlike with other derivatives, these functions and their derivatives 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!

• Note with the derivative of $\arctan x$ that $\left(1+x^2\right)$ is the same as $\left(x^2+1\right)$

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

• For $y=\arcsin x$

  • Rewrite, $\sin y=x$

  • Differentiate implicitly, $\cos y\frac{\mathrm{d}y}{\mathrm{d}x}=1$

  • Rearrange, $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\cos y}$

  • Rewrite using the identity ${\cos }^{2}y\equiv 1-{\sin }^{2}y$, $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\sqrt{1-{\sin }^{2}y}}$

  • And $\sin y=x$, therefore $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\sqrt{1-x^2}}$

• Similarly, for $y=\arccos x$

  • $\cos y=x$

  • $-\sin y\frac{\mathrm{d}y}{\mathrm{d}x}=1$

  • $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{\sin y}$

  • $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{\sqrt{1-{\cos }^{2}y}}$

  • $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{\sqrt{1-x^2}}$

• For proving the derivative of $y=\arctan x$ , see the Worked Example

• Notice how the derivative of $y=\arcsin x$ is positive but the derivative of $y=\arccos x$ is negative

  • 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 derivatives of inverse trig functions look like with linear functions of x?

• For linear functions of the form $ax+b$

• $f\left(x\right)=\arcsin \left(ax+b\right)$

  • $f^{\text{'}}\left(x\right)=\frac{a}{\sqrt{1-{(ax+b)}^2}}$

• $f\left(x\right)=\arccos \left(ax+b\right)$

  • $f^{\text{'}}\left(x\right)=-\frac{a}{\sqrt{1-{(ax+b)}^2}}$

• $f\left(x\right)=\arctan \left(ax+b\right)$

  • $f^{\text{'}}\left(x\right)=\frac{a}{1+{(ax+b)}^2}$

• 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