Proofs of the formulas of the derivatives of inverse trigonometric functions are presented along with several other examples involving sums, products, and quotients of functions. Another method to find the derivative of inverse functions is also included and may be used.
Proofs and Formulas for Inverse Trig Derivatives
1. Derivative of \( y = \arcsin(x) \)
Let \( y = \arcsin(x) \), which may be written as \( x = \sin(y) \).
Differentiate both sides with respect to \( x \):
\[ \dfrac{dx}{dx} = \dfrac{d(\sin(y))}{dx} \implies 1 = \cos(y) \dfrac{dy}{dx} \implies \dfrac{dy}{dx} = \dfrac{1}{\cos y} \]Using the trigonometric identity \( \cos(y) = \sqrt{1 - \sin^2(y)} = \sqrt{1 - x^2} \), we obtain:
\[ \mathbf{\dfrac{d(\arcsin(x))}{dx} = \dfrac{1}{\sqrt{1 - x^2}}} \]2. Derivative of \( \arccos(x) \)
Let \( y = \arccos(x) \), which may be written as \( x = \cos(y) \).
Differentiating both sides with respect to \( x \):
\[ 1 = -\sin(y) \dfrac{dy}{dx} \implies \dfrac{dy}{dx} = -\dfrac{1}{\sin(y)} \]Since \( \sin(y) = \sqrt{1 - \cos^2(y)} = \sqrt{1 - x^2} \), we obtain:
\[ \mathbf{\dfrac{d(\arccos(x))}{dx} = -\dfrac{1}{\sqrt{1 - x^2}}} \]3. Derivative of \( \arctan(x) \)
Let \( y = \arctan(x) \), which may be written as \( x = \tan(y) \).
Differentiate both sides with respect to \( x \):
\[ 1 = \sec^2(y) \dfrac{dy}{dx} \implies \dfrac{dy}{dx} = \dfrac{1}{\sec^2(y)} = \cos^2(y) \]Using the identity \( \cos^2(y) = \dfrac{1}{1 + \tan^2(y)} = \dfrac{1}{1 + x^2} \), we obtain:
\[ \mathbf{\dfrac{d(\arctan(x))}{dx} = \dfrac{1}{1 + x^2}} \]4. Derivative of \( \text{arccot}(x) \)
Let \( y = \text{arccot}(x) \), which may be written as \( x = \cot(y) \).
Differentiate both sides with respect to \( x \):
\[ 1 = -\csc^2(y) \dfrac{dy}{dx} \implies \dfrac{dy}{dx} = -\dfrac{1}{\csc^2(y)} = -\sin^2(y) \]Since \( \sin^2(y) = \dfrac{1}{1 + \cot^2(y)} = \dfrac{1}{1 + x^2} \), we obtain:
\[ \mathbf{\dfrac{d(\text{arccot}(x))}{dx} = -\dfrac{1}{1 + x^2}} \]5. Derivative of \( \text{arcsec}(x) \)
Let \( y = \text{arcsec}(x) \), which may be written as \( x = \sec(y) \).
Differentiate both sides with respect to \( x \):
\[ 1 = \sec(y) \tan(y) \dfrac{dy}{dx} \implies \dfrac{dy}{dx} = \dfrac{1}{\sec(y) \tan(y)} \]Using \( \tan y = \sqrt{\sec^2 y - 1} = \sqrt{x^2 - 1} \), we obtain:
\[ \mathbf{\dfrac{d(\text{arcsec}(x))}{dx} = \dfrac{1}{x \sqrt{x^2 - 1}}} \]6. Derivative of \( \text{arccsc}(x) \)
Let \( y = \text{arccsc}(x) \), which may be written as \( x = \csc(y) \).
Differentiate both sides with respect to \( x \):
\[ 1 = -\csc(y) \cot(y) \dfrac{dy}{dx} \implies \dfrac{dy}{dx} = -\dfrac{1}{\csc(y) \cot(y)} \]Using \( \cot y = \sqrt{\csc^2 y - 1} = \sqrt{x^2 - 1} \), we obtain:
\[ \mathbf{\dfrac{d(\text{arccsc}(x))}{dx} = -\dfrac{1}{x \sqrt{x^2 - 1}}} \]Examples with Solutions
Example 1
Find the first derivative of:
\[ f(x) = x \arcsin x \]Solution to Example 1
Let \( h(x) = x \) and \( g(x) = \arcsin x \). Function \( f \) is considered as the product of functions \( h \) and \( g \): \( f(x) = h(x)g(x) \).
Use the product rule of differentiation: \( f'(x) = h(x)g'(x) + g(x)h'(x) \):
\[ f'(x) = x \left(\dfrac{1}{\sqrt{1 - x^2}}\right) + \arcsin x \cdot 1 \] \[ = \dfrac{x}{\sqrt{1 - x^2}} + \arcsin x \]Example 2
Find the first derivative of:
\[ f(x) = \arctan x + x^2 \]Solution to Example 2
Let \( g(x) = \arctan x \) and \( h(x) = x^2 \). Function \( f \) may be considered as the sum of functions \( g \) and \( h \): \( f(x) = g(x) + h(x) \). Hence we use the sum rule, \( f'(x) = g'(x) + h'(x) \):
\[ f'(x) = \dfrac{1}{1 + x^2} + 2x \] \[ = \dfrac{1 + 2x(1 + x^2)}{1 + x^2} = \dfrac{2x^3 + 2x + 1}{1 + x^2} \]Example 3
Find the first derivative of:
\[ f(x) = \arcsin (2x + 2) \]Solution to Example 3
Let \( u(x) = 2x + 2 \), so function \( f \) is the composition \( f(x) = \arcsin(u(x)) \). Use the chain rule, \( f'(x) = \dfrac{du}{dx} \cdot \dfrac{d(\arcsin(u))}{du} \):
\[ f'(x) = 2 \left(\dfrac{1}{\sqrt{1 - u^2}}\right) \]Substitute \( u = 2x + 2 \):
\[ = \dfrac{2}{\sqrt{1 - (2x + 2)^2}} \]