Question

Differentiate the following funetions: \(u=x(1-x)^{4}\)

Short answer

Question: Find the derivative of the function \(u = x(1-x)^{4}\). Answer: \(u'(x) = (1-x)^{3}(1-5x)\).

Step-by-step solution

Identify the functions

We have the function \(u(x) = x(1-x)^{4}\). We will treat it as a product of two functions: - \(f(x) = x\) - \(g(x) = (1-x)^{4}\)

Differentiate f(x)

The derivative of \(f(x) = x\) with respect to x is: \(f'(x) = 1\)

Differentiate g(x)

To find the derivative of \(g(x) = (1-x)^{4}\), we will use the chain rule: \(g'(x) = \frac{d}{dx}((1-x)^{4}) = 4(1-x)^{3}(-1)\) Simplifying, we get: \(g'(x) = -4(1-x)^{3}\)

Apply the Product Rule

Now we will apply the product rule for differentiation. With \(f'(x) = 1\) and \(g'(x) = -4(1-x)^{3}\), we have: \(u'(x) = f'(x)g(x) + f(x)g'(x) = 1(1-x)^{4} + x(-4)(1-x)^{3}\)

Simplify the result

Finally, we simplify the expression for the derivative: \(u'(x) =(1-x)^{3}[(1-x) - 4x]\) And that's our final answer: \(u'(x) = (1-x)^{3}(1-5x)\).

Key concepts

Product Rule

In calculus, the product rule is a valuable tool used for differentiating products of two functions. It's particularly useful when dealing with functions like the one in our exercise, where we have a product of two terms,

• \(f(x) = x\)
• \(g(x) = (1-x)^4\)

The product rule states that if you have a function \(u(x) = f(x) \, g(x)\), then the derivative \(u'(x)\) is given by:\[ u'(x) = f'(x)g(x) + f(x)g'(x) \]This formula helps break down the differentiation process into manageable steps by treating each function component, \(f(x)\) and \(g(x)\), separately. When applying the product rule, make sure to differentiate both functions independently before substituting them back into the formula. In our exercise, this yields the derivative:\[ u'(x) = 1(1-x)^{4} + x(-4)(1-x)^{3} \]Applying the product rule is like dismantling a complex object into simpler parts, solving them, and then putting them back together for a complete solution.

Chain Rule

The chain rule is another essential technique in calculus differentiation. It helps differentiate composite functions, which is when one function is nested inside another, like \(g(x) = (1-x)^4\) in our example. The chain rule formula is:\[ \frac{d}{dx}[h(g(x))] = h'(g(x)) \, g'(x) \]To use the chain rule, identify the outer function \(h\) and the inner function \(g\). Differentiate the outer function, leaving the inner part unchanged, and then multiply by the derivative of the inner function.

• In "\((1-x)^4\)", the outer function is "\((\cdot)^4\)" and the inner function is "\((1-x)\)".

Differentiating the outer function results in \(4(1-x)^3\), and multiplying by the derivative of the inner function \((-1)\), gives us:\[ g'(x) = -4(1-x)^3 \]This step-by-step process ensures that no part of the composite function is left undifferentiated.

Calculus Differentiation

Calculus differentiation forms the backbone of mathematical analysis in change and rates, such as speed and acceleration. The primary goal in differentiation is to find the derivative of functions, which indicates the function's rate of change. In our example, we applied both the product rule and the chain rule to find \(u'(x)\). These rules are foundational methods in doing calculus differentiation effectively.Engage in the practice of identifying the parts of the function:

• Differentiate each part using the appropriate rule.
• Simplify the resulting expression for an accurate result.

Understanding these core concepts is crucial because it allows you to approach complex functions with structured strategies, breaking them down into simpler components. With repeated practice, these rules become second nature, making differentiation a much smoother process. Our final result for the derivative, \(u'(x) = (1-x)^{3}(1-5x)\), illustrates the concise clarity calculus differentiation brings to function analysis.