Question

Compute the derivative of the given function. $$h(t)=\sin ^{-1}(2 t)$$

Short answer

The derivative of \( h(t) = \sin^{-1}(2t) \) is \( \frac{2}{\sqrt{1-4t^2}} \).

Step-by-step solution

Identify the Function Type

The given function is \(h(t) = \sin^{-1}(2t)\), which is the inverse sine (arcsine) function. We need to differentiate it with respect to \(t\).

Recall the Derivative Formula for Arcsine Function

The derivative of \(\sin^{-1}(x)\) with respect to \(x\) is \(\frac{1}{\sqrt{1-x^2}}\). Because our function is \(\sin^{-1}(2t)\), we'll need to apply the chain rule.

Apply the Chain Rule

Using the chain rule, the derivative of \(\sin^{-1}(2t)\) with respect to \(t\) is \(\frac{d}{dt}[\sin^{-1}(u)]\cdot \frac{du}{dt}\), where \(u = 2t\).

Differentiate \(u\) with Respect to \(t\)

Differentiate \(u = 2t\). The derivative is \(\frac{du}{dt} = 2\).

Differentiate the Arcsine Function

The derivative of \(\sin^{-1}(u)\) with respect to \(u\) is \(\frac{1}{\sqrt{1-u^2}}\). Substitute \(u=2t\) into this formula to get \(\frac{1}{\sqrt{1-(2t)^2}} = \frac{1}{\sqrt{1-4t^2}}\).

Combine Results Using Chain Rule

Combine the derivatives using the chain rule: \(\frac{dh}{dt} = \frac{1}{\sqrt{1-4t^2}} \cdot 2\).

Simplify the Expression

Simplify the expression for the derivative: \(\frac{dh}{dt} = \frac{2}{\sqrt{1-4t^2}}\).

Key concepts

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. Imagine a function within another function—whenever you have this setup, you'll need the chain rule. In simple terms, the chain rule tells us how to differentiate the outer function while also accounting for the inner function.
Suppose you have a composite function defined as \(h(t) = f(g(t))\). The chain rule states that the derivative of this function is the product of the derivative of the outer function and the derivative of the inner function. In mathematical terms:

• \(\frac{dh}{dt} = \frac{df}{dg} \cdot \frac{dg}{dt}\)

Here, \(\frac{df}{dg}\) is the derivative of the outer function, and \(\frac{dg}{dt}\) is the derivative of the inner function. This rule is powerful and widely applicable, allowing you to handle situations where functions are embedded inside other functions.

Arcsine Differentiation

Arcsine differentiation involves differentiating the inverse sine function. When you see \(\sin^{-1}(x)\), you are dealing with an arcsine function. The derivative of arcsine, \(\frac{d}{dx}[\sin^{-1}(x)]\), is a unique formula given by:

• \(\frac{1}{\sqrt{1-x^2}}\)

This derivative involves a square root and is valid only within the domain of the arcsine function, i.e., when \(|x| \leq 1\).
This constraint ensures the expression inside the square root remains non-negative, fulfilling the mathematical requirements for real number computation. When the function involves a transformation like in the problem \(\sin^{-1}(2t)\), you'd use the chain rule to first differentiate the inner function.

Inverse Trigonometric Functions

Inverse trigonometric functions are the functions that "undo" the traditional trigonometric functions like sine, cosine, and tangent. There are six primary inverse trigonometric functions:

• Arcsine, \(\sin^{-1}(x)\)
• Arccosine, \(\cos^{-1}(x)\)
• Arctangent, \(\tan^{-1}(x)\)
• Arccosecant, \(\csc^{-1}(x)\)
• Arcsecant, \(\sec^{-1}(x)\)
• Arccotangent, \(\cot^{-1}(x)\)

For the context of differentiation, each inverse function has a specific formula governing its derivative. Differentiating involves understanding the range and domain, since the inverse functions can only "work" within certain limits.
For example, the derivative of each inverse trig function is connected to its original function and often incorporates a square root, displaying the critical role of ensuring the argument remains within a valid range.