Question

Compute the derivative of the given function. $$h(x)=\frac{\sin ^{-1} x}{\cos ^{-1} x}$$

Short answer

The derivative of the function is \( h'(x) = \frac{\frac{\pi}{2}}{\sqrt{1-x^2} \cdot [\cos^{-1}(x)]^2} \).

Step-by-step solution

Identify the Functions

The function given is \( h(x) = \frac{\sin^{-1}(x)}{\cos^{-1}(x)} \), where \( \sin^{-1}(x) \) and \( \cos^{-1}(x) \) are the inverse sine and inverse cosine functions, respectively.

Apply the Quotient Rule

Recall that the derivative of a quotient \( \frac{u(x)}{v(x)} \) is given by \[ h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \] For our function, identify \( u(x) = \sin^{-1}(x) \) and \( v(x) = \cos^{-1}(x) \).

Find Derivatives of Inverse Trig Functions

The derivative of \( \sin^{-1}(x) \) is \( u'(x) = \frac{1}{\sqrt{1-x^2}} \). The derivative of \( \cos^{-1}(x) \) is \( v'(x) = -\frac{1}{\sqrt{1-x^2}} \).

Substitute into the Quotient Rule

Substitute \( u(x) = \sin^{-1}(x) \), \( v(x) = \cos^{-1}(x) \), \( u'(x) = \frac{1}{\sqrt{1-x^2}} \), and \( v'(x) = -\frac{1}{\sqrt{1-x^2}} \) into the quotient rule formula, resulting in:\[h'(x) = \frac{\frac{1}{\sqrt{1-x^2}}\cos^{-1}(x) - \sin^{-1}(x)(-\frac{1}{\sqrt{1-x^2}})}{[\cos^{-1}(x)]^2}\]

Simplify the Derivative

Combine the terms in the numerator:\[ h'(x) = \frac{\cos^{-1}(x) + \sin^{-1}(x)}{\sqrt{1-x^2} \cdot [\cos^{-1}(x)]^2} \] Since \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \), simplify further to:\[ h'(x) = \frac{\frac{\pi}{2}}{\sqrt{1-x^2} \cdot [\cos^{-1}(x)]^2} \]

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse operations of the standard trigonometric functions, such as sine and cosine. They allow us to find the angle that corresponds to a given trigonometric ratio.
The main types include:

• Inverse Sine, denoted as \(\sin^{-1}(x)\) or \(\arcsin(x)\)
• Inverse Cosine, denoted as \(\cos^{-1}(x)\) or \(\arccos(x)\)
• Inverse Tangent, denoted as \(\tan^{-1}(x)\) or \(\arctan(x)\)

These functions are particularly useful when we want to determine angles from known ratios.
For example, if \(\sin(\theta) = x\), then \(\theta = \sin^{-1}(x)\). Similarly, these identities help when dealing with calculus problems, such as differentiation of trigonometric expressions.

Quotient Rule

The quotient rule is a technique used in calculus to find the derivative of a function that is the ratio of two differentiable functions.
For any functions \(u(x)\) and \(v(x)\) where \(v(x) eq 0\), the derivative of their quotient \(\frac{u(x)}{v(x)}\) is expressed as:
\[h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\]
This rule is crucial when dealing with complex fractions in calculus.

• It helps in maintaining accuracy while differentiating quotients compared to other methods like product rule or the power rule.
• Applying this rule involves differentiating the numerator and the denominator separately, then using the formula.

This ensures that we get the most simplified and accurate form of the derivative of quotient functions.

Differentiation

Differentiation is the process used in calculus to determine how a function changes at any given point.
In simple terms, it's about finding the rate of change or the slope of a function.
When differentiating a function, one aims to find its derivative, which represents an equation for this rate of change.

• Differentiation formulas are used for basic functions like power functions, trigonometric functions, and exponentials.
• The quotient rule, product rule, and chain rule are specific techniques for differentiating different types of functions.

Understanding differentiation is fundamental to more advanced calculus concepts and is widely applied across physics, engineering, and economics to solve practical problems.

Inverse Sine

The inverse sine function, \(\sin^{-1}(x)\), is used to find the angle whose sine is \(x\).
This function is restricted to the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\) to maintain its uniqueness and functionality as a proper inverse function.
When differentiating \(\sin^{-1}(x)\), you'll find its derivative to be:\[ \frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}} \]
This expression arises because the rate of change of \(\sin^{-1}(x)\) is dependent on the value of \(x\), which is between -1 and 1.

• This derivative is particularly useful when differentiating composite functions involving inverse sine.
• Always ensure \(x\) remains within the domain to use this derivative appropriately.

Inverse Cosine

Similarly, the inverse cosine function, denoted as \(\cos^{-1}(x)\), helps in determining the angle whose cosine is \(x\).
The range for \(\cos^{-1}(x)\) is \([0, \pi]\), allowing it to be well-defined as an inverse function.
The derivative of the inverse cosine function is:\[ \frac{d}{dx} \cos^{-1}(x) = -\frac{1}{\sqrt{1-x^2}} \]
Like the inverse sine derivative, it is defined for \(-1 \leq x \leq 1\). The negative sign indicates the decreasing nature of the cosine function.

• This derivative is essential when calculating derivatives of complex functions involving inverse cosine.
• Always check the domain of \(x\) to ensure proper application of this derivative.