Question

Evaluate the following derivatives. \(f(t)=2 \tanh ^{-1} \sqrt{t}\)

Short answer

Answer: The derivative of \(f(t) = 2\text{arctanh}(\sqrt{t})\) is \(f'(t) = \frac{1}{(1 - t)\sqrt{t}}\).

Step-by-step solution

Applying the chain rule

First, we need to apply the chain rule. Recall that the chain rule states that if we have a composition of functions \(f(g(t))\), the derivative is given by \(f'(g(t)) \cdot g'(t)\). In this case, our outer function is \(2\text{arctanh}(x)\) and our inner function is \(\sqrt{t}\). Applying the chain rule, we get: $$ f'(t) = \frac{d}{dt} [2\text{arctanh}(\sqrt{t})] = 2 \cdot \frac{d}{d(\sqrt{t})} [\text{arctanh}(\sqrt{t})] \cdot \frac{d}{dt} [\sqrt{t}] $$ Now, we need to find the derivatives \(\frac{d}{d(\sqrt{t})} [\text{arctanh}(\sqrt{t})]\) and \(\frac{d}{dt} [\sqrt{t}]\).

Finding the derivative of \(\text{arctanh}(x)\)

Recall that the derivative of \(\text{arctanh}(x)\) is given by: $$ \frac{d}{dx} [\text{arctanh}(x)] = \frac{1}{1 - x^2} $$ Now we can find the derivative of \(\text{arctanh}(\sqrt{t})\) with respect to \(\sqrt{t}\): $$ \frac{d}{d(\sqrt{t})} [\text{arctanh}(\sqrt{t})] = \frac{1}{1 - (\sqrt{t})^2} = \frac{1}{1 - t} $$

Finding the derivative of \(\sqrt{t}\)

Now, let's find the derivative of \(\sqrt{t}\) with respect to \(t\): $$ \frac{d}{dt} [\sqrt{t}] = \frac{d}{dt} [t^{1/2}] = \frac{1}{2}t^{-1/2} $$

Combining the derivatives

Now we have all the required derivatives and can combine them to find \(f'(t)\): $$ f'(t) = 2 \cdot \frac{1}{1 - t} \cdot \frac{1}{2}t^{-1/2} = \frac{1}{(1 - t)\sqrt{t}} $$ Therefore, the derivative of the given function is: $$ f'(t) = \frac{1}{(1 - t)\sqrt{t}} $$

Key concepts

Chain Rule

The chain rule is a fundamental technique in calculus for differentiating compositions of functions. Consider a function composed of two or more functions, like \( f(g(t)) \). Here, the chain rule states that the derivative is the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function.

• Begin by identifying your outer and inner functions. In our problem, the outer function is \(2\text{arctanh}(x)\), and the inner function is \(\sqrt{t}\).
• The derivative of \(f(t) = 2 \text{arctanh}(\sqrt{t})\) uses the chain rule: \( f'(t) = 2 \cdot [ \text{arctanh}'(\sqrt{t}) ] \cdot [ (\sqrt{t})' ]\).
• This method allows for effectively dealing with composite functions, especially when different types of functions are involved.

By dissecting each part of the function, you begin to appreciate the essence of the rule: handling the 'links' in compositions to differentiate them seamlessly.
Take your time to practice recognizing inner and outer functions as this is key in efficiently applying the chain rule.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are counterparts to the regular trigonometric functions but for hyperbolic functions like sine and cosine. Specifically, they inversely relate to hyperbolic functions.

• Examples include \(\text{arsinh}(x)\), \(\text{arcosh}(x)\), and our case, \(\text{arctanh}(x)\).
• They are especially crucial in calculus due to their properties in modeling and differentiating complex phenomena.
• Because these functions have log-based derivations, they show up in calculus often during integration and differentiation exercises.

Understanding these functions involves recognizing how they reverse their counterpart hyperbolic functions and knowing essential differentiation formulas. Practicing these will enhance your skill in handling calculus problems that feature inverse hyperbolic functions.
Start by studying typical inverse hyperbolic expressions and their graphical interpretations to get a solid grasp.

Derivative of Arctanh

The derivative of the inverse hyperbolic tangent function, denoted as \(\text{arctanh}(x)\), is an essential formula in calculus. When differentiating \(\text{arctanh}(x)\), you employ the fact:

The derivative is \( \frac{1}{1 - x^2} \), which emerges from the inverse relationship and the hyperbolic identity.

This derivative is pivotal when encountering the function within complex derivatives. As demonstrated, this formula allows analysis and derivation in nested functions.

This property simplifies the derivative process in our example where \(\sqrt{t}\) replaces \(x\). Recognizing the derivative's form assists in calculating and applying the chain rule appropriately.
Memorizing these special derivatives can significantly speed up solving inverse hyperbolic calculus problems.

Power Rule

The power rule is a straightforward yet powerful differentiation tool useful for functions of the form \(x^n\). It states that for \(f(x) = x^n\), the derivative \(f'(x) = n \cdot x^{n-1}\). This rule simplifies taking derivatives of power functions.

• In our context, we use it to differentiate \(\sqrt{t}\), which is equivalent to \(t^{1/2}\).
• Applying the power rule gives \(\frac{1}{2}t^{-1/2}\) as its derivative.

The simplicity of this rule lies in its application to any real number exponent, ensuring ease in handling routine calculus tasks.
As it forms the fundamental base for numerous derivations, practice with a variety of exponents will strengthen your grasp of calculus functionalities.