Problem 45 - Calculus and its applications

Question

Find \(\frac{d y}{d u}, \frac{d u}{d x},\) and \(\frac{d y}{d x}\) $$ y=\frac{15}{u^{3}} \text { and } u=2 x+1 $$

Step-by-Step Solution

Answer

\(\frac{d y}{d u} = -\frac{45}{u^4}\), \(\frac{d u}{d x} = 2\), and \(\frac{d y}{d x} = -\frac{90}{(2x+1)^4}\).

Step 1: Differentiate y with respect to u

To find \( \frac{d y}{d u} \), we first differentiate the function \( y = \frac{15}{u^3} \) with respect to \( u \). Using the power rule and chain rule, we have:\[\frac{d y}{d u} = -3 \times \frac{15}{u^{4}} = -\frac{45}{u^{4}}.\]

Step 2: Differentiate u with respect to x

Now, we will differentiate \( u = 2x + 1 \) with respect to \( x \). The derivative of \( 2x + 1 \) is simply the constant from the linear term, so:\[\frac{d u}{d x} = 2.\]

Step 3: Apply Chain Rule to find dy/dx

To find \( \frac{d y}{d x} \), we apply the chain rule, which states that: \[\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}.\]Now substitute the derivatives we calculated:\[\frac{d y}{d x} = \left(-\frac{45}{u^{4}} \right) \times 2 = -\frac{90}{u^{4}}.\]Now, since \( u = 2x + 1 \), substitute back into the expression:\[\frac{d y}{d x} = -\frac{90}{(2x+1)^{4}}.\]

Key Concepts

Chain Rule

The chain rule is an essential tool in differentiation, particularly when dealing with composite functions. A composite function is a function made up of two or more functions, such as \( y = f(g(x)) \). In such cases, the chain rule helps us find the derivative of the composite function by breaking it down into simpler parts.

The chain rule states that:

• To differentiate a composite function, multiply the derivative of the outer function by the derivative of the inner function.
• Mathematically, for \( y = f(g(x)) \), the derivative \( \frac{d y}{d x} = f'(g(x)) \times g'(x) \).

In our example, where \( y = \frac{15}{u^3} \) and \( u = 2x + 1 \), the chain rule is applied as follows:

• First, find \( \frac{d y}{d u} \) using \( y \) in terms of \( u \).
• Next, find \( \frac{d u}{d x} \) by differentiating \( u \) with respect to \( x \).
• Finally, multiply the two derivatives to get \( \frac{d y}{d x} \).

Power Rule

The power rule is handy when taking derivatives of polynomial and rational functions. It simplifies finding the derivative of expressions involving powers of a variable.

The power rule states:

• For a function \( f(x) = x^n \), the derivative \( f'(x) = n \times x^{n-1} \).

This rule applies not only to simple polynomials but also to handle expressions like \( \frac{1}{u^3} \). By rewriting, \( \frac{1}{u^3} \) can also be expressed as \( u^{-3} \).

In our exercise:

• When finding \( \frac{d y}{d u} \), \( y = \frac{15}{u^3} = 15 \times u^{-3} \). Using the power rule: the derivative is \(-3 \times 15 \times u^{-4} = -\frac{45}{u^4}\).

Derivative of Polynomial Functions

Derivatives of polynomial functions play a crucial role in calculus as they describe how a polynomial function changes. When dealing with expressions like \( u = 2x + 1 \), these tend to be linear polynomials.

Here’s how to differentiate a simple linear polynomial:

• For \( u(x) = ax + b \), the derivative \( \frac{d u}{d x} = a \).

The derivative is constant since it represents the slope of the line formed by the polynomial. For our given function, \( u = 2x + 1 \), the derivative is quite straightforward:

• The slope, or derivative, \( \frac{d u}{d x} \), is 2, meaning the line rises 2 units for every 1 unit it moves horizontally.

Using this derivative, when combined with other rules like the chain rule, we can find the overall rate of change in more complex functions such as \( y = \frac{15}{u^3} \).