Derivative of Inverse Trigonometric Functions

Consider the equation:

$$ \sin^{-1}(2x) + \frac{\pi}{4} = \tan^{-1}\left( \frac{x}{\sqrt{1 - x^2}} \right), \quad 0 < x < 1 $$

Solution: The solution satisfies: $$ x = \frac{-1}{\sqrt{10 - 4\sqrt{2}}} $$

Prove the following derivatives:

a. $\frac{d}{dx}[\cos^{-1} x] = -\frac{1}{\sqrt{1 - x^2}}$

b. $\frac{d}{dx}[\tan^{-1} x] = \frac{1}{1 + x^2}$

c. $\frac{d}{dx}[\csc^{-1} x] = -\frac{1}{|x|\sqrt{x^2 - 1}}$

Find the first derivative of the following functions:

a. $y = \sin^{-1}(3x)$

b. $y = \cos^{-1}\left( \frac{x+1}{2} \right), \quad -1 < x < 1$

c. $y = \sin^{-1}\left( \frac{1}{x} \right), \quad x > 0$

d. $y = \sec^{-1}(x^2)$

e. $y = \tan^{-1}(e^{2x})$

f. $y = \ln(x^2) \sec^{-1}(4x^3)$

Differentiate the following functions:

a. $y = x + \sin^{-1}(e^{2x})$

b. $y = \tan^{-1}(x^2) \csc^{-1}(\ln x)$

c. $y = \frac{\cos^{-1}(2x)}{3x - e^{2x}}$

d. $y = \frac{e^{3x} \sin^{-1}(5x)}{\ln(x^2) \tan x}$

Show that: $$ \cosh x + \sinh x = e^{x} $$

Prove that: $$ \tanh x = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} $$ and $$ \operatorname{sech} x = \frac{2}{e^{x} + e^{-x}} $$

Hence, verify the identity: $$ \operatorname{sech}^{2} x = 1 - \tanh^{2} x $$

Source: FAC1004 Tutorial 7 25-26.pdf