Question

Differentiate the following funetions: \(u=x(a+b x)^{n}\)

Short answer

Question: Find the derivative of the function \(u(x) = x(a + bx)^n\) with respect to \(x\). Answer: \(u'(x) = (a + bx)^n + bn x(a + bx)^{n-1}\)

Step-by-step solution

Identify the functions

Firstly, we identify the two functions that we need to differentiate: 1. \(f(x) = x\) 2. \(g(x) = (a + bx)^n\)

Differentiate f(x) and g(x) individually

Now, we differentiate each function with respect to \(x\): For \(f(x) = x\), we have: \(f'(x) = 1\) For \(g(x) = (a + bx)^n\), we apply the chain rule (let \(h(x) = a + bx\), and \(g(x) = h(x)^n\)): \(g'(x) = n h(x)^{n-1} \cdot h'(x)\) Then, differentiate \(h(x) = a + bx\) with respect to \(x\): \(h'(x) = b\) Now, substitute the value of \(h'(x)\) back into the expression of \(g'(x)\): \(g'(x) = n (a + bx)^{n-1} \cdot b\)

Apply the product rule

We now apply the product rule to differentiate the function \(u(x) = f(x) \cdot g(x)\): \(u'(x) = f'(x)g(x) + f(x)g'(x)\)

Substitute the values and simplify

Substitute the values of \(f'(x)\), \(g(x)\), and \(g'(x)\) that we found in Steps 2 and 3 into the expression for \(u'(x)\): \(u'(x) = 1 \cdot (a + bx)^n + x \cdot n (a + bx)^{n-1} \cdot b\) Now, we can simplify the expression: \(u'(x) = (a + bx)^n + bn x(a + bx)^{n-1}\) So, the derivative of the given function is: \(u'(x) = (a + bx)^n + bn x(a + bx)^{n-1}\)

Key concepts

Product Rule Differentiation

The product rule is a fundamental technique in calculus used for finding the derivative of the product of two functions. When you have a function that is the product of two other functions, like
e.g., if you have a function in the form of
\( u(x) = f(x) \cdot g(x) \), rather than just a single-term function, the product rule enables you to find the derivative of this composite function.

The formula for the product rule is given by:
\[ u'(x) = f'(x)g(x) + f(x)g'(x) \]

This says that the derivative of the product of two functions \( f(x) \) and \( g(x) \) is the derivative of the first function \( f(x) \) times the second function \( g(x) \) plus the first function times the derivative of the second function \( g'(x) \). It's essential to keep each function and its derivative correctly paired.

In practice, you will first identify the two standalone functions that are multiplied together, differentiate each function independently, and then use the product rule formula for the product.

Example in Action

In the given exercise, \( u(x) = x(a+b x)^{n} \), \( f(x) = x \) and \( g(x) = (a + bx)^n \) were identified as the functions being multiplied. Differentiating \( f(x) \) gave \( f'(x) = 1 \), and the derivative of \( g(x) \) was found by applying the chain rule, leading to the final expression of the product of these two functions.

Chain Rule Differentiation

The chain rule is a technique in calculus used to differentiate composite functions. This rule is vital when dealing with nested functions – functions that are composed of one function inside another.

The chain rule states that the derivative of a composite function
\( g(f(x)) \) is the derivative of the outer function evaluated at the inner function times the derivative of the inner function:
\[ g'(f(x)) \cdot f'(x) \]

To apply the chain rule, we often denote the inner function as \( h(x) \).

Application to the Exercise

In our exercise, we looked at the function \( g(x) = (a + bx)^n \), treating \( h(x) = a + bx \) as the inside function and \( g(x) = h(x)^n \) as the outside function. Following the chain rule,
\[ g'(x) = n \cdot h(x)^{n-1} \cdot h'(x)\], where \( h'(x) = b \), the derivative of the inner function. This concept enables us to differentiate functions that are not plainly laid out as simple polynomials but have layers to work through.

Power Rule Differentiation

The power rule simplifies the process of differentiating polynomials where each term is a power of \( x \). It states that if you have a function of the form
\( f(x) = x^n \), then its derivative is
\[ f'(x) = n \cdot x^{n-1} \].

This rule stands as one of the most straightforward differentiation rules to remember and apply. For any exponent \( n \), be it a positive or negative integer or even a rational number, applying the power rule will give the correct derivative straight away.

In the case of our exercise, the power rule is used after applying the chain rule to differentiate \( g(x) = (a + bx)^n \). The exponent \( n \) in the function comes down to multiply ahead of the function, and then we decrease the power by one. On its own, the power rule might seem too trivial, but it's especially powerful when combined with other rules like the chain and product rules.

Calculus Step by Step Solutions

Calculus can often be intimidating, but breaking it down into step by step solutions can make it much more accessible. Providing calculus solutions in such a structured manner helps learners not only reach the answer but also understand the underlying principles.

A step by step solution typically involves identifying the functions involved, differentiating each function according to the respective rules (such as the product rule, chain rule, and power rule), and then combining these derivatives in a meaningful way according to the original problem statement.

For our purposes, the exercise to differentiate \( u(x) = x(a+b x)^{n} \) was tackled by first identifying the two functions that make up the product, differentiating each separately, and finally applying the product rule. The detailed process helps students follow the logic of calculus, see the intermediate steps, and better grasp how multiple differentiation rules may be applied together to find the derivative of a more complex function. In such solutions, clarity is key, and each step must be explained so that the students can follow the logical progression of the problem-solving process.