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Chapter 9: Problem 47

Find the derivative of the function. \(f(x)=(3 x+1)^{4}\left(x^{2}-x+1\right)^{3}\)

Short Answer

Expert verified
The derivative of the function \(f(x)=(3 x+1)^{4}\left(x^{2}-x+1\right)^{3}\) is: \(\frac{df}{dx} = 3(3x+1)^4(x^2-x+1)^2(2x-1) + 12(3x+1)^3(x^2-x+1)^3\)

Step by step solution

01

Recall the Product Rule and Chain Rule formulas

The Product Rule states that if we have a function in the form of \(u(x)v(x)\), the derivative is given by \(\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\). The Chain Rule states that if we have a composite function in the form of \(y = f(g(x))\), the derivative is given by \(\frac{dy}{dx} = \frac{df}{dg} \frac{dg}{dx}\).
02

Apply the Product Rule

Let \(u(x)=(3x+1)^4\) and \(v(x)=(x^2-x+1)^3\). We want to find the derivative of \(f(x)=u(x)v(x)\). Using the Product Rule, we have: \(\frac{df}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\) Now, we need to find the derivatives \(\frac{du}{dx}\) and \(\frac{dv}{dx}\).
03

Apply the Chain Rule to find du/dx

To find \(\frac{du}{dx}\), let \(g(x) = 3x+1\), so \(u(x) = g(x)^4\). Applying the Chain Rule, we get: \(\frac{du}{dx} = \frac{d(g^4)}{dg}\cdot\frac{dg}{dx}\) Using the Power Rule for derivatives, \(\frac{d(g^4)}{dg} = 4g^{4-1}\), and \(\frac{dg}{dx} = 3\). So, we get: \(\frac{du}{dx} = 4g^3\cdot 3 = 12(3x+1)^3\)
04

Apply the Chain Rule to find dv/dx

To find \(\frac{dv}{dx}\), let \(h(x) = x^2-x+1\), so \(v(x) = h(x)^3\). Applying the Chain Rule, we get: \(\frac{dv}{dx} = \frac{d(h^3)}{dh}\cdot\frac{dh}{dx}\) Using the Power Rule for derivatives, \(\frac{d(h^3)}{dh} = 3h^{3-1}\), and \(\frac{dh}{dx} = 2x-1\). So, we get: \(\frac{dv}{dx} = 3h^2\cdot (2x-1) = 3(x^2-x+1)^2(2x-1)\)
05

Plug in the derivatives into the Product Rule formula

We now substitute the calculated derivatives back into the Product Rule formula: \(\frac{df}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} = (3x+1)^4\cdot\left[3(x^2-x+1)^2(2x-1)\right] + (x^2-x+1)^3\cdot\left[12(3x+1)^3\right]\) Combining these terms, we get the final derivative of the function: \(\frac{df}{dx} = 3(3x+1)^4(x^2-x+1)^2(2x-1) + 12(3x+1)^3(x^2-x+1)^3\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

The Product Rule
When the function we're dealing with is a multiplication of two separate functions, like in the exercise \(f(x)=(3 x+1)^{4}(x^{2}-x+1)^{3}\), we can't just individually take derivatives of each part and multiply them together. Instead, we use the Product Rule for differentiation. The rule is a formula that helps us find the derivative of such products efficiently. It states that if you have two differentiable functions \(u(x)\) and \(v(x)\), the derivative of their product \(uv\) with respect to \(x\) is \(u'v + uv'\), where \(u'\) is the derivative of \(u\) with respect to \(x\), and \(v'\) is the derivative of \(v\).

The Chain Rule
Functions often come in layers, with one function nested inside another. This is where the Chain Rule comes into play. Imagine you鈥檙e taking off a set of Russian dolls; that's similar to how we use the Chain Rule鈥攚e differentiate from the outside in. This is critical when dealing with composite functions, as seen in our exercise where \(u(x)=(3x+1)^4\) and \(v(x)=(x^2-x+1)^3\).

First, you differentiate the outer function as if the inner portion were a simple variable. Then, you multiply it by the derivative of the inner function. Mathematically, if \(y = f(g(x))\), then \(\frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}\). The Chain Rule is used in our exercise to find the derivatives of \(u\) and \(v\) with respect to \(x\), considering \(g(x) = 3x+1\) and \(h(x) = x^2-x+1\), respectively.
The Power Rule
Raising functions to a power is a common operation, and the Power Rule simplifies finding their derivatives. The rule is quite simple: if you have a function \(g(x) = x^n\), where \(n\) is a real number, the derivative of that function with respect to \(x\) is \(ng(x)^{n-1}\). This comes in handy when you can write your function as something to the power of \(n\), just like in our exercise with \(g(x)^4\) and \(h(x)^3\).

By applying the Power Rule, you can easily find the derivatives that are needed for both the Chain Rule and the Product Rule. This is the essence of what makes calculus work鈥攖hese rules can be stacked and used together to find the derivatives of even the most complex functions. And by understanding and applying these rules step by step, as shown in the solution for the given exercise, differentiation becomes a much more manageable task.

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Most popular questions from this chapter

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