91Ó°ÊÓ

One of the essential rules in calculus differentiation is the power rule. It is used to differentiate functions of the form \(x^n\), where \(n\) is a real number.

The power rule states:

\(\frac{d}{dx} x^n = nx^{n-1} \).

This rule tells us that to find the derivative of \(x^n\), we multiply by the exponent \(n\) and then subtract one from the exponent in the resulting term.

For example, if we have \(f(x) = x^5\), we differentiate as follows:

• Use the power rule: \(f'(x) = 5x^{5-1}\)



• Simplify the exponent: \(f'(x) = 5x^4\)

This process makes differentiation straightforward. It simplifies finding the slope of a curve defined by a polynomial function.

In calculus, the product rule is used to find the derivative of the product of two functions. It is very useful when dealing with more complex functions that are the product of two simpler functions.

The product rule is stated as follows:

• If we have two functions \(u(x)\) and \(v(x)\), the derivative of their product \(uv\) is:


\[(uv)' = u'v + uv'\]


• Here, \(u'\) is the derivative of \(u(x)\), and \(v'\) is the derivative of \(v(x)\).

Let's break down how to use the product rule with an example:

Suppose \(y = x^3 \times x^8\). We can find the derivative using the product rule. Define \(u = x^3\) and \(v = x^8\). Then, we need:

• \(u' = 3x^2\)



• \(v' = 8x^7\)



• Apply the product rule: \(y' = (x^3)' \times x^8 + x^3 \times (x^8)'\)



• So, \(y' = 3x^2 \times x^8 + x^3 \times 8x^7 = 3x^{10} + 8x^{10}\)

Simplify to get:

\(y' = 11x^{10}\)

This methodical approach allows us to differentiate products of functions efficiently.

Simplifying exponents is crucial in calculus to make differentiation easier. It involves applying basic exponent rules to combine and simplify terms.

Let's consider the expression given in the exercise, \(y = x^3 \times x^8\). To simplify, we use the rule:

• When multiplying like bases, add their exponents: \(a^m \times a^n = a^{m+n}\).

Applying this to the expression gives:

• \(y = x^3 \times x^8\)



• Add the exponents: \(y = x^{3+8}\)



• Simplify to: \(y = x^{11}\)

This step makes the function easier to work with, particularly when differentiating. Instead of handling the product of two terms, we now only need to differentiate a single term.

Taking the simplified form and using the power rule, the derivative is straightforward:

• \(y = x^{11}\)



• Using the power rule: \(y' = 11x^{10}\)

By simplifying exponents early, we streamline the process for further differentiation, saving time and reducing potential errors.