Question

Calculate the derivative of the following functions. $$y=\sqrt[3]{x^{2}+9}$$

Short answer

Answer: The derivative of the given function is $$\frac{dy}{dx} = \frac{2x}{3(x^{2}+9)^{\frac{2}{3}}}$$

Step-by-step solution

Identify the outer and inner functions

The given function can be represented as a composition of two functions, with an outer function and an inner function. Here, the outer function is $$f(u) = \sqrt[3]{u}$$ and the inner function is $$g(x) = x^{2}+9$$. So, $$y = f(g(x))$$.

Differentiate the outer function with respect to u

We need to differentiate the outer function, $$f(u)=\sqrt[3]{u}$$. Using the power rule of differentiation, we have $$f'(u) = \frac{1}{3}u^{-\frac{2}{3}}$$.

Differentiate the inner function with respect to x

Now, we need to differentiate the inner function, $$g(x) = x^{2}+9$$. Using the power rule of differentiation, we have $$g'(x) = 2x$$.

Apply the Chain Rule

Now that we have found the derivatives of the outer and inner functions, we can apply the Chain Rule to find the derivative of the given function. The Chain Rule states that $$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$$.

Substitute the values and simplify

Substitute $$f'(u)$$ and $$g'(x)$$ into the Chain Rule equation: $$\frac{dy}{dx} = \frac{1}{3}(x^{2}+9)^{-\frac{2}{3}}\cdot 2x$$ Multiply the terms to simplify: $$\frac{dy}{dx} = \frac{2x}{3(x^{2}+9)^{\frac{2}{3}}}$$ Hence, the derivative of the given function is $$\frac{dy}{dx} = \frac{2x}{3(x^{2}+9)^{\frac{2}{3}}}$$

Key concepts

Chain Rule Differentiation

Grasping the chain rule is vital for calculus students as it allows for the differentiation of composite functions. When a function is composed of one function inside of another, the chain rule enables us to find its derivative by multiplying the derivative of the outer function by the derivative of the inner function.

For example, consider the function in our exercise \( y = \sqrt[3]{x^{2}+9} \). Here, the chain rule comes into play because the function consists of an inner function \( g(x) = x^2 + 9 \) and an outer function, which is the cubed root. When differentiating complex functions like this, identify the 'inner layer' and 'outer layer' of the function first, differentiate them separately, and then apply the chain rule to combine their derivatives. This concept is one of the cornerstones of calculus and is widely used across different applications, from physics to economics.

Power Rule of Differentiation

When dealing with polynomials or any terms with an exponent, the power rule is a quick method to differentiate. Given a function \( f(x) = x^n \), the power rule tells us that the derivative \( f'(x) \) is \( nx^{n-1} \).

In our given function \( y = \sqrt[3]{x^{2}+9} = (x^2+9)^{\frac{1}{3}} \), we apply the power rule to both the inner function \( g'(x) = 2x^1 \) and the outer function after it's transformed to \( f(u) = u^{\frac{1}{3}} \) yielding \( f'(u) = \frac{1}{3}u^{-\frac{2}{3}} \). This rule greatly simplifies the process of finding derivatives and underscores why it is a fundamental tool in calculus.

Composite Function Derivative

The concept of a composite function derivative emerges when we have one function nested within another, and it's a direct application of the chain rule. Composite functions are often written as \( f(g(x)) \), where \( g \) is the inner function and \( f \) is the outer function.

In the exercise, we have \( y = f(g(x)) \), where \( f(u) = \sqrt[3]{u} \) and \( g(x) = x^2+9 \). After finding derivatives of both \( f \) and \( g \) separately, we combine them using the chain rule to find the derivative of the entire function \( y \). This illustrates how understanding the composite function derivative is crucial for handling more complex functions in calculus.

Calculus Problem Solving

Problem-solving in calculus often involves a series of strategic steps. It starts with understanding the problem, identifying what is required, and recognizing which rules and concepts apply. From there, breaking down a function into manageable parts—like outer and inner functions—can simplify the process.

In our example, we start by identifying the inner and outer functions, proceed to differentiate each part, and then combine our results using the chain rule. Problem-solving skills also include simplifying expressions and checking the correctness of the answer. Mastery over these strategies can significantly enhance a student's ability to tackle a wide range of calculus problems with ease and confidence.