Problem 48 - Calculus Early Transcendentals

Question

Find the derivatives of the following functions. $$f(t)=2 \tanh ^{-1} \sqrt{t}$$

Step-by-Step Solution

Answer

Answer: The derivative of the function is $$f'(t) = \frac{1}{(1 - t)\sqrt{t}}$$.

Step 1: Let's first identify the inner function in our given function $$f(t) = 2 \tanh^{-1} \sqrt{t}$$. The inner function is $$u = g(t) = \sqrt{t}$$. #Step 2: Differentiate the inner function#

Now, we'll find the derivative of the inner function $$g(t) = \sqrt{t}$$. Using the power rule, we have: $$g'(t) = \frac{d}{dt} \sqrt{t} = \frac{d}{dt} t^{\frac{1}{2}} = \frac{1}{2} t^{-\frac{1}{2}}$$ #Step 3: Differentiate the outer function#

Step 2: Next, we'll find the derivative of the outer function $$F(u) = 2 \tanh^{-1}(u)$$. Using the formula for the derivative of the inverse hyperbolic tangent function, we get: $$F'(u) = 2 \times \frac{1}{1 - u^2}$$ #Step 4: Apply the chain rule#

Now that we have both derivatives, we can apply the chain rule to find the derivative of our original function. According to the chain rule, the derivative of our function is given by: $$f'(t) = F'(g(t)) * g'(t)$$ #Step 5: Plug the derivatives into the chain rule formula#

Step 3: Finally, we can substitute our derivatives into the chain rule formula: $$f'(t) = \left(2 \times \frac{1}{1 - (\sqrt{t})^2}\right) \times \left(\frac{1}{2} t^{-\frac{1}{2}}\right)$$ #Step 6: Simplify the derivative#

Let's simplify the expression we obtained in the previous step: $$f'(t) = \left(2 \times \frac{1}{1 - t}\right) \times \left(\frac{1}{2} t^{-\frac{1}{2}}\right) = \frac{1}{(1 - t)\sqrt{t}}$$ So, the derivative of our original function is: $$f'(t) = \frac{1}{(1 - t)\sqrt{t}}$$

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus that helps us differentiate composite functions. Imagine you have a function made up of two simpler functions, embedded one inside the other. The Chain Rule allows you to break this down and differentiate step by step. Consider the function \(f(t) = 2 \tanh^{-1} \sqrt{t}\) mentioned in the exercise.

• Firstly, identify the inner and outer functions. Here \(u = \sqrt{t}\) is the inner function and \(F(u) = 2 \tanh^{-1}(u)\) is the outer function.
• Differentiate the inner function \(g(t) = \sqrt{t}\), resulting in \(g'(t) = \frac{1}{2} t^{-\frac{1}{2}}\).
• Then, differentiate the outer function with respect to \(u\): \(F'(u) = \frac{2}{1 - u^2}\).

Now, apply the Chain Rule: the derivative of the composite function is \[f'(t) = F'(g(t)) \times g'(t)\].
In our example, this gives \(f'(t) = \left(\frac{2}{1-t}\right) \cdot \left(\frac{1}{2} t^{-\frac{1}{2}}\right)\).

By putting everything together and simplifying, the final derivative becomes \( \frac{1}{(1 - t)\sqrt{t}} \).

Using the Chain Rule efficiently can simplify the process of finding derivatives of complex functions.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are extensions of regular hyperbolic functions and play a crucial role in calculus, often appearing in problems involving integrals and derivatives. They are akin to their trigonometric cousins, allowing for the inverse computation of hyperbolic angles.

• The inverse hyperbolic tangent, denoted as \(\tanh^{-1}(x)\), is one of these inverse functions. It is the inverse of the hyperbolic tangent function \(\tanh(x)\).
• These functions are named similarly to inverse trigonometric functions and are used primarily in solving equations involving hyperbolic functions.

The derivative of inverse hyperbolic tangent is especially important. It can be derived using implicit differentiation or understanding its basic relationship, where:
\[\frac{d}{dx}\left(\tanh^{-1}(x)\right) = \frac{1}{1-x^2}\].

Understanding inverse hyperbolic functions and their derivatives contributes to a broader comprehension of calculus, especially when tackling integrals and derivatives of more complex functions, like in our exercise.

Derivative of Inverse Hyperbolic Tangent

The derivative of the inverse hyperbolic tangent function \(\tanh^{-1}(x)\) is a central part of the problem we are solving. Understanding its derivation makes it easier to tackle similar questions in the future.

• The formula for the derivative of \(\tanh^{-1}(x)\) is: \(\frac{d}{dx} \tanh^{-1}(x) = \frac{1}{1 - x^2}\).
• This formula comes from recognizing the rate of change of the function, with respect to its input \(x\).
• In this exercise, \(\tanh^{-1}(u)\) is scaled by a factor of \(2\), so the derivative becomes \(2 \times \frac{1}{1 - u^2}\)."

Understanding this derivative is key, as it allows us to correctly apply the Chain Rule and solve for the derivative of complex composite functions. In our example, using this derivative alongside the Chain Rule helps to reach the simplified final answer of \( \frac{1}{(1-t)\sqrt{t}} \).

Mastering the derivatives of inverse hyperbolic functions like \(\tanh^{-1}(x)\) is essential for excelling in calculus.