Question

Finding a Derivative In Exercises \(7-34,\) find the derivative of the function. $$ f(x)=x^{2}(x-2)^{4} $$

Short answer

The derivative of the function \(f(x)=x^{2}(x-2)^{4}\) is \(f'(x) = 6x^2(x-2)^3\).

Step-by-step solution

Identify the Product Rule

The Product Rule in differentiation is: If we have two functions ‘u’ and ‘v’, then the derivative of the product uv is given by \(u(v') + v(u')\). Here we have our two functions as \(u = x^{2}\) and \(v = (x-2)^{4}\).

Differentiate u and v

Calculate derivatives for u and v independently. The derivative of \(x^{2}\) (u) is \(2x\) and the derivative of \((x-2)^{4}\) (v) requires the application of Chain Rule. The Chain Rule is given by \(dy/dx = dy/du * du/dx\) where y is a function of u which is again function of x. Here, for v, if we consider \((x-2)\) as 'u' and '(u)^4' as 'y', then the derivative \(dy/du = 4u^{3}\) and \(du/dx = 1\). Thus, the derivative of v, \(v' = dy/dx = dy/du * du/dx = 4(u^{3})*(1) = 4(x-2)^3\).

Apply the Product Rule

Now, combine the derivatives using the Product Rule. That is, the derivative of \(f(x)=x^{2}(x-2)^{4}\) is \(u'v + vu' = 2x*(x-2)^4 + x^2*4(x-2)^3\). This can be simplified further.

Simplify the Expression

Given expression can be simplified by factoring. Factor out an \(x(x-2)^3\) to get the final derivative. So, the derivative of \(f(x)=x^{2}(x-2)^{4}\) is \(f'(x)=x(x-2)^3[2(x-2) + 4x] = 2x^2(x-2)^3 +4x^3(x-2)^3 = 6x^2(x-2)^3\).

Key concepts

Product Rule

The product rule is a fundamental concept in differentiation utilized when we wish to find the derivative of a product of two functions. Imagine you have two functions, \(u(x)\) and \(v(x)\), which are multiplied together to create a new function, \(f(x) = u(x) \cdot v(x)\).
Stating the rule simply, the derivative of this product is given by:

• \(f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\).

Essentially, we take the derivative of the first function and multiply it by the second function, then take the derivative of the second function and multiply it by the first. Add these two products together for the final result!
In our specific problem, \(u(x) = x^2\) and \(v(x) = (x-2)^4\). By applying the product rule, we differentiate each part separately and combine them as described.

Chain Rule

The chain rule is another pivotal rule in calculus used to differentiate composite functions. A composite function occurs when one function is nested inside another, such as \(y = (g(x))^n\). The chain rule formula is:

• \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

In simpler terms, first, you find the derivative of the outer function at the inner function, then multiply it by the derivative of the inner function.
In our exercise, to find the derivative of \((x-2)^4\), we treat \(x-2\) as an inner function and \(u^4\) as the outer function. Applying the chain rule, we first differentiate \((u^4)\) with respect to \(u\), obtaining \(4u^3\), then multiply by the derivative of \(u = x-2\), which is \(1\). Therefore, the derivative of \((x-2)^4\) becomes \(4(x-2)^3\).
The chain rule is indispensable for handling compounded functions in calculus problems.

Differentiation

Differentiation is the process of finding a derivative, and it represents the rate at which a function is changing at any given point.
It is the fundamental concept in calculus which helps us to understand the behavior of functions. Differentiating a function answers the question: how does the output of a function change as the input changes?

• For example, the derivative of a constant function is zero.
• Power functions follow the rule \(\frac{d}{dx}(x^n) = nx^{n-1}\).

If you have functions multiplied or composed with each other, as seen in our exercise, rules like the product and chain rules are necessary. In our exercise, differentiating \(x^2\) gives \(2x\), clearly adhering to the power rule. This simple differentiation step is a building block combined with others to solve complex equations.

Factorization

Factorization often comes at the end of a differentiation problem. It is the process of breaking down complex expressions into simpler components, which can make the expression easier to analyze and simplify.
After applying the product and chain rules, we often get a long and complex derivative. Factorization helps us tidy this up. In our problem, after differentiating, we simplified:

• \(2x(x-2)^4 + x^2(4(x-2)^3)\).

By spotting common factors, in this case, \(x(x-2)^3\), we pulled it out:
\(2x^2(x-2)^3 + 4x^3(x-2)^3 = x(x-2)^3[2(x-2) + 4x]\)
This reveals hidden structure and simplifies the expression greatly, often making mathematical relationships clearer and simpler to interpret. Factorization is thus a generic tool but essential in streamlining differentiation results.