Problem 35 - Single Variable Calculus

Question

Find the derivative of the function. $$ f(x)=\left(1+\sin ^{2} 3 x\right)^{2 / 3} $$

Step-by-Step Solution

Answer

The short answer for the derivative of the function \(f(x)=\left(1+\sin ^{2} 3 x\right)^{2 / 3}\) is: \[ f'(x)=\frac{4\sin{3x}\cos{3x}}{\sqrt[3]{\left(1+\sin^2{3x}\right)}} \]

Step 1: Identify the outer and inner functions

The given function can be rewritten as \(f(x) = g(h(x))\), where - The outer function: \(g(u)=(u)^{2/3}\) - The inner function: \(h(x)=1+\sin^2{3x}\) To differentiate, we need to find both \(g'(u)\) and \(h'(x)\) and use the chain rule formula: \(f'(x) = g'(h(x))\cdot h'(x)\).

Step 2: Differentiate the outer function

To differentiate the outer function, we will apply the power rule, which states that for any function \(y=u^n\), the derivative is \(y'=nu^{n-1}\). In our case, \(u^n=u^{2/3}\), so the derivative is: \(g'(u) = \frac{2}{3}u^{\frac{2}{3}-1} = \frac{2}{3}u^{-\frac{1}{3}}\)

Step 3: Differentiate the inner function

To differentiate the inner function, \(h(x) = 1+\sin^2{3x}\), we will need to apply both the chain rule and the sine function's derivative. The derivative of a constant (1) is zero, so we only need to differentiate \(\sin^2{3x}\). To do this, let's rewrite it as \((\sin{3x})^2\), so we have an outer function (\(v^2\)) and an inner function (\(\sin{3x}\)). Differentiate the outer function (\(v^2\)): \(2v\) Differentiate the inner function (\(\sin{3x}\)): \(\cos{3x}\cdot 3 = 3\cos{3x}\) Using the chain rule, the derivative of \((\sin{3x})^2\) is: \((2v)\cdot(3\cos{3x}) = 6\sin{3x}\cos{3x}\) Now we have the derivative of the inner function: \(h'(x) = 0 + 6\sin{3x}\cos{3x}\)

Step 4: Apply the chain rule to find the derivative of the original function

We now have all the required derivatives to apply the chain rule. Recall that the chain rule formula is: \(f'(x) = g'(h(x))\cdot h'(x)\). So, the derivative of the given function is: \(f'(x) = \left(\frac{2}{3}\left(1+\sin^2{3x}\right)^{-\frac{1}{3}}\right) \cdot \left(6\sin{3x}\cos{3x}\right)\) Simplify the expression to get the final answer: \(f'(x)=\frac{12\sin{3x}\cos{3x}}{3\sqrt[3]{\left(1+\sin^2{3x}\right)}}\) \(f'(x)=\frac{4\sin{3x}\cos{3x}}{\sqrt[3]{\left(1+\sin^2{3x}\right)}}\)

Key Concepts

Chain Rule

The Chain Rule is a technique in calculus used to differentiate composite functions. If you have a function that is composed of two or more functions nested within each other, the Chain Rule is your tool of choice for finding its derivative. Suppose you have a composite function given by \(f(x) = g(h(x))\). This means "\(f\)" is the result of "\(g\)" acting upon "\(h(x)\)".
To apply the Chain Rule, you need to:

• Identify the outer function \(g(u)\) and the inner function \(h(x)\).
• Find the derivative of the outer function: \(g'(u)\).
• Find the derivative of the inner function: \(h'(x)\).
• Multiply these derivatives together: \(f'(x) = g'(h(x)) \cdot h'(x)\).

In our example with \(f(x)=\left(1+\sin^2 3x\right)^{2/3}\),

• the outer function is \(\left(u\right)^{2/3}\)
• the inner function is \(1+\sin^2{3x}\)

We used the Chain Rule to blend these derivatives, giving us the final derivative \(f'(x)\). Remember, the key to mastering the Chain Rule is to practice identifying and differentiating both the inner and outer functions correctly.

Power Rule

The Power Rule is one of the simplest rules for finding derivatives, and it's a cornerstone of calculus. It states that if you have a function of the form \(y = u^n\), the derivative \(y'\) will be \(n \cdot u^{n-1}\).
This rule applies to both basic polynomial functions and situations where the base, \(u\), is itself a more complex expression.
In our problem, we applied the Power Rule to the outer function \(g(u) = u^{2/3}\) as part of the Chain Rule. By differentiating it:

• we found \(g'(u) = \frac{2}{3}u^{-1/3}\).

This application of the Power Rule is a critical step when differentiating composite functions because it manages the exponent portion effectively. So, remember that any time you see a power on a variable or expression, the Power Rule will likely be your method of choice.

Trigonometric Derivatives

Trigonometric derivatives are vital when you have functions involving sine, cosine, and other trigonometric expressions. In calculus, these derivatives follow specific rules:

• The derivative of \(\sin(a)\) is \(\cos(a)\).
• The derivative of \(\cos(a)\) is \(-\sin(a)\).
• These derivatives may involve applying the Chain Rule when the angle itself is a function, such as \(3x\) in our example.

In our exercise, the inner function, \(h(x) = 1 + \sin^2(3x)\), involved a trigonometric expression \((\sin(3x))^2\).
To differentiate, we needed to:

• Apply the Power Rule to \((\sin{3x})^2\).
• Use trigonometric derivatives for \(\sin{3x}\).

Thus, we explored these steps and found that \(\sin^2(3x)\) led us to use its derivative in the form \(6\sin{3x}\cos{3x}\), utilizing \(\sin{2a} = 2\sin{a}\cos{a}\) identity.
This shows the interconnectedness of trigonometry and calculus, where understanding fundamental identities is essential to differentiating trigonometric functions effectively.