Question

The function \(y(x)\) is defined by \(y(x)=\left(1+x^{m}\right)^{n}\). (a) Use the chain rule to show that the first derivative of \(y\) is \(n m x^{m-1}\left(1+x^{m}\right)^{n-1}\). (b) The binomial expansion (see section \(1.5\) ) of \((1+z)^{n}\) is $$ (1+z)^{n}=1+n z+\frac{n(n-1)}{2 !} z^{2}+\cdots+\frac{n(n-1) \cdots(n-r+1)}{r !} z^{r}+\cdots $$ Keeping only the terms of zeroth and first order in \(d x\), apply this result twice to derive result (a) from first principles. (c) Expand \(y\) in a series of powers of \(x\) before differentiating term by term. Show that the result is the series obtained by expanding the answer given for \(d y / d x\) in (a).

Short answer

\(y' = nm x^{m-1} (1 + x^m)^{n-1}\). This is confirmed using binomial expansion and term-by-term differentiation.

Step-by-step solution

Use the chain rule (part a)

Given the function \(y(x) = (1 + x^m)^n\), we need to find the first derivative using the chain rule. Let \(u = 1 + x^m\), then \(y = u^n\). The chain rule states that \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

Differentiate \(y\) with respect to \(u\)

Using the power rule, the derivative of \(y = u^n\) with respect to \(u\) is \(\frac{dy}{du} = n u^{n-1}\).

Differentiate \(u\) with respect to \(x\)

Given \(u = 1 + x^m\), differentiate it with respect to \(x\): \( \frac{du}{dx} = m x^{m-1}\).

Apply the chain rule

Combine the results from Steps 2 and 3: \( \frac{dy}{dx} = n u^{n-1} \cdot m x^{m-1} \). Substitute \(u\) back in: \( \frac{dy}{dx} = n (1 + x^m)^{n-1} \cdot m x^{m-1} \). This simplifies to \( \frac{dy}{dx} = nm x^{m-1} (1 + x^m)^{n-1} \).

Apply the binomial expansion (part b)

The binomial expansion is given by: \((1 + z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \cdots\). First apply this to \( y = (1 + x^m)^n \) with \( z = x^m \). We get: \(y \approx 1 + n x^m\).

Compare differential terms

When expanding in terms of \(dx\), we focus on the terms up to the first order. When differentiating both sides, we get: \(dy \approx n x^{m-1} dx\). This confirms the derivative.

Expand in a power series (part c)

Write \(y\) as a power series: \( (1 + x^m)^n \). Expand using a Taylor series: \(y = 1 + n x^m + \frac{n(n-1)}{2!}(x^m)^2 + \cdots\).

Differentiate term-by-term

Differentiate each term with respect to \(x\): \( \frac{d}{dx}[1 + n x^m + \frac{n(n-1)}{2!}(x^m)^2 + \cdots] \). This becomes: \( 0 + nm x^{m-1} + k x^{k(m-1)} + \cdots \). This matches the given form of \(d y / d x\).

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. If you have a function that can be expressed as a composition of two functions, say \( y(x) = f(g(x)) \), the chain rule helps you find the derivative of \( y \) with respect to \( x \). To do this, you need to first differentiate the outer function \( f \) with respect to the inner function \( g \), and then multiply it by the derivative of the inner function \( g \) with respect to \( x \).

In this exercise, we applied the chain rule to the function \( y(x) = (1 + x^m)^n \). Here's how:

• First, let \( u = 1 + x^m \), defining the inner function.
• Now, the outer function becomes \( y = u^n \).
• By differentiating \( y \) with respect to \( u \), we get \( \frac{dy}{du} = n u^{n-1} \).
• Next, differentiate \( u \) with respect to \( x \): \( \frac{du}{dx} = m x^{m-1} \).

Finally, multiply these results: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = n u^{n-1} \cdot m x^{m-1} \).

Substitute \( u \) back in to align with the original function, which results in \( \frac{dy}{dx} = nm x^{m-1} (1 + x^m)^{n-1} \).

Binomial Expansion

The binomial expansion allows us to expand expressions like \( (1 + z)^n \) into a series. This is especially useful for approximating expressions when \( z \) is a small number.

The binomial theorem states that:
\[ (1+z)^{n} = 1 + nz + \frac{n(n-1)}{2!}z^{2} + \cdots + \frac{n(n-1) \cdots(n-r+1)}{r!} z^{r} + \cdots \]

In our exercise, we used this expansion by setting \( z = x^m \). By keeping only the zeroth and first order terms, we get an approximation: \( y \approx 1 + nx^m \).

When we differentiate this with respect to \( x \):

• The zeroth order term (1) vanishes since its derivative is 0.
• For the first order term \( nx^m \), the derivative is \( nm x^{m-1} \).

Thus, we obtain \( \frac{dy}{dx} \approx nm x^{m-1} \). This confirms the first step derivative result.

Taylor Series

The Taylor series allows us to represent a function as an infinite sum of terms calculated from the values of its derivatives at a single point. When expanding a function using a Taylor series, we can approximate complex functions with simpler polynomial terms.

In this exercise, we expanded the function \( y(x) = (1 + x^m)^n \) as a power series. This means we write it as:
\[ y(x) = 1 + nx^m + \frac{n(n-1)}{2!}(x^m)^2 + \cdots \]

Next, we differentiated this series term by term:

• The constant term (1) becomes 0 when differentiated.
• The first term \( nx^m \) gives \( nm x^{m-1} \).
• Continuing this process for higher-order terms, we get:

\( \frac{d}{dx} \left[ \frac{n(n-1)}{2!}(x^m)^2 \right] = n(n-1) mx^{m-1} \cdot x^m \).

Combining these results, we find that the series form of \( \frac{dy}{dx} \) matches the derived form \( nm x^{m-1}(1 + x^m)^{n-1} \). This demonstrates the consistency of the differentiation process whether we use a series expansion or direct methods.