Problem 62 - Single Variable Calculus

Question

Find the derivative of the function. $$ f(x)=\tan ^{-1} \sqrt{3 x+1} $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = \tan^{-1} \sqrt{3x + 1}\) is \(f'(x) = \frac{3}{4x + 2\sqrt{3x + 1}}\).

Step 1: Identify the Inner and Outer Functions

Let's first analyze the given function to find the inner and outer functions involved. In this case, we can see that the function is a composition of \(\tan^{-1}x\) and \(\sqrt{3x+1}\). So, we have: - Outer function: \(g(x) = \tan^{-1}x\) - Inner function: \(h(x) = \sqrt{3x+1}\)

Step 2: Find the Derivatives of the Inner and Outer Functions

Now, we will find the derivatives of the outer and inner functions with respect to \(x\). For the outer function, we have \(g(x) = \tan^{-1}x\). The derivative of \(\tan^{-1}x\) is \(\frac{1}{1+x^2}\). So, we get: - \(\frac{dg}{dx} = \frac{1}{1+x^2}\) For the inner function, we have \(h(x) = \sqrt{3x+1}\), which can also be written as \((3x+1)^{1/2}\). To find the derivative of the inner function, we will apply the power rule combined with the chain rule. We get: - \(\frac{dh}{dx} = \frac{1}{2}(3x+1)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x+1}}\)

Step 3: Apply the Chain Rule to the Function

Next, we will apply the chain rule to find the derivative of the given function, \(f(x) = g(h(x))\). Using the chain rule, we have: \[\frac{df}{dx} = \frac{dg}{dh} \cdot \frac{dh}{dx}\] Substituting the values of \(\frac{dg}{dx}\) and \(\frac{dh}{dx}\) that we found in Step 2, we get: \[\frac{df}{dx} = \frac{1}{1+\left(\sqrt{3x+1}\right)^2} \cdot \frac{3}{2\sqrt{3x+1}}\]

Step 4: Simplify the Expression

Finally, we will simplify the expression for \(\frac{df}{dx}\): \[\frac{df}{dx} = \frac{1}{1+3x+1} \cdot \frac{3}{2\sqrt{3x+1}} = \frac{3}{4x+2\sqrt{3x+1}}\] So, the derivative of the given function is: \[\boxed{\frac{df}{dx} = \frac{3}{4x+2\sqrt{3x+1}}}\]

Key Concepts

Chain Rule

The chain rule is an essential concept in calculus used for finding the derivative of composite functions. A composite function can be thought of as a function within another function, which means you have an inner function and an outer function.

When one function is nested inside another, to find the derivative, you differentiate the outer function first and then multiply it by the derivative of the inner function. Symbolically, if a function is given by \(f(x) = g(h(x))\), its derivative \(f'(x)\) is found via the chain rule as \(f'(x) = g'(h(x)) \cdot h'(x)\).

In the context of our exercise, the chain rule allows us to systematically take the derivative of \( f(x)=\tan ^{-1} \sqrt{3 x+1} \) by first taking the derivative of the outer function \( \tan^{-1}x \) and then multiplying it by the derivative of the inner function \( \sqrt{3x+1} \). It simplifies the complex process of differentiation by breaking it down into manageable steps.

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative when functions are given in an implicit form rather than the standard explicit form \(y = f(x)\). Sometimes establishing a relationship directly between \(x\) and \(y\) is not feasible, and instead, they are related by an equation that involves both \(x\) and \(y\).

With implicit differentiation, you differentiate both sides of the equation with respect to \(x\) and then solve for \(\frac{dy}{dx}\). Since the derivative of \(y\) with respect to \(x\) is not directly obtainable, you treat \(y\) as an implicit function of \(x\) during differentiation.

While implicit differentiation is not directly applied in our initial exercise, it is instrumental when dealing with inverse functions such as \( \tan^{-1}x \), where the relationship between \(y\) and \(x\) may be implicitly defined.

Inverse Functions

Inverse functions essentially reverse the action of a given function. If a function \( f \) takes an input \( x \) and produces an output \( y \) such that \( y = f(x) \), then its inverse function \( f^{-1} \) will take \( y \) as an input to give an output \( x \) such that \( x = f^{-1}(y) \).

One property of inverse functions is that their derivatives are reciprocal to each other under certain conditions. For example, if you have a function \( f \) and its inverse \( f^{-1} \) then \( (f^{-1})'(y) = \frac{1}{f'(x)} \), where \( x = f^{-1}(y) \). The trigonometric function in our exercise, \( \tan^{-1}x \) is the inverse function of the tangent function, which is why its derivative is reciprocal to the derivative of the tangent function.

Power Rule

The power rule is one of the most straightforward rules for differentiation. It is used when we need to find the derivative of a function that is a power of \(x\). The power rule states that if you have a function of the form \( f(x) = x^n \), where \( n \) is any real number, the derivative of the function is \( f'(x) = n \cdot x^{n-1} \).

In practice, this means you bring down the exponent as a coefficient and subtract one from the exponent. The power rule becomes a powerful tool when combined with the chain rule, as seen in our exercise. The inner function \( h(x) = \sqrt{3x+1} \) is rewritten as \( (3x+1)^{1/2} \), and by applying the power rule, we find its derivative, which later multiplies with the derivative of the outer function according to the chain rule.