Question

In Exercises \(1-8,\) find the derivative of \(y\) with respect to the appropriate variable. $$y=\sin ^{-1}(1-t)$$

Short answer

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

Step-by-step solution

Identify the outer and inner function.

We first identify the outer function and the inner function. In this case, the outer function is the inverse sine function, and the inner function is \(1-t\). So, \(f(t) = \sin^{-1}(t)\) and \(g(t) = 1-t\). We need to find the derivative of \(y = f(g(t))\) with respect to \(t\).

Find the derivatives of the outer and inner function.

Next, we find the derivatives of the outer and inner function using their respective derivative rules. The derivative of \(\sin^{-1}(t)\) is \(\frac{1}{\sqrt{1-t^2}}\), and the derivative of \(1-t\) is \(-1\). We denote these as \(f'(t)\) and \(g'(t)\) respectively.

Apply the chain rule.

Now, we apply the chain rule to find the derivative of the composite function. According to the chain rule, the derivative of a composite function is given by the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function, i.e., \((f\circ g)'(t) = f'(g(t)) \cdot g'(t)\). Substituting the derivatives and functions from Steps 1 and 2, we find that the derivative of \(y = \sin^{-1}(1-t)\) is \((f\circ g)'(t) = \frac{-1}{\sqrt{1-(1-t)^2}} = \frac{-1}{\sqrt{t(2-t)}}\)

Simplifying the expression.

Finally, we simplify the fraction to get the final derivative. We do not need to simplify in this case as our derivative is already in the simplest form. Therefore, the derivative of \(y = \sin^{-1}(1-t)\) is \(\frac{-1}{\sqrt{t(2-t)}}\)

Key concepts

Inverse Trigonometric Functions Derivatives

Understanding the derivatives of inverse trigonometric functions is crucial for calculus students, especially when dealing with complex composite functions. Inverse trigonometric functions, also known as arc-functions, provide the angles whose trigonometric functions are known values. For instance, the arcsine function, written as \(\sin^{-1}\), yields the angle whose sine value is given.

Here's an essential fact: the derivative of \(\sin^{-1}(x)\) is \(\frac{1}{\sqrt{1-x^2}}\). Why is this important? Because when you're working with composite functions—where the arcsine function is of some function of x, not just x itself—you'll need to apply the chain rule to find the complete derivative, taking into account both the outer and inner functions' derivatives.

It is highly recommended to familiarize yourself with each inverse trigonometric function's derivative. Such as for \(\tan^{-1}(x)\), the derivative is \(\frac{1}{1+x^2}\), or for \(\cos^{-1}(x)\), it is \(-\frac{1}{\sqrt{1-x^2}}\). These derivatives are fundamental building blocks when it comes to solving calculus problems involving inverse functions.

Composite Function Differentiation

Composite functions are like nesting dolls of mathematics; you've got a function tucked within another function. Differentiation of composite functions is not as straightforward as simple functions, which is where our topic, the chain rule, comes into play. When you differentiate a composite function, you're taking the derivative of the 'outer' function with respect to the 'inner' function, and then multiplying by the derivative of the 'inner' function with respect to its own variable.

To get the hang of differentiating composite functions, it's beneficial to practice identifying the inner and outer functions in various examples. For instance, if you have a function like \(h(x) = (3x+2)^4\), the outer function is \(u^4\) where \(u = 3x + 2\), which is your inner function. Recognizing these components is the first step before you can apply the chain rule correctly. Always remember, smooth differentiation requires comfort with this recognition process.

Applying the Chain Rule

The chain rule is the bread and butter for differentiating composite functions. In simpler terms, it's a method to find the derivative of a function based on another function. Think of it as a way to unlink the chain of composite functions to treat each link—each function—separately.

Here's the gist of it: If you have two functions, say \(f(x)\) and \(g(x)\), where one is composed with the other to form \(h(x) = f(g(x))\), the chain rule states that the derivative of \(h(x)\) with respect to \(x\) is \(h'(x) = f'(g(x))\cdot g'(x)\). This approach allows you to find the derivative of complex functions by breaking them down into their simpler parts, then using their derivatives accordingly.

To apply the chain rule effectively, start by finding the derivatives of the outer and inner functions separately before multiplying them together. As seen in the exercise, we first differentiate the outer function, incorporated with the inner function, and then apply the derivative of the inner function. This multi-step process may appear daunting, but with practice, it becomes a natural part of solving calculus problems.