Problem 9 - Basic Engineering Mathematics

Question

Find the differential coefficient of \(y=\frac{2}{5} x^{3}-\frac{4}{x^{3}}+4 \sqrt{x^{5}}+7\)

Step-by-Step Solution

Answer

The differential coefficient is \(\frac{6}{5}x^2 + 12x^{-4} + 10x^{3/2}\).

Step 1: Identify the Terms for Differentiation

The given function is \(y = \frac{2}{5}x^3 - \frac{4}{x^3} + 4\sqrt{x^5} + 7\). We need to identify each term before differentiating: \(\frac{2}{5}x^3\), \(-\frac{4}{x^3}\), and \(4\sqrt{x^5}\). Remember that constants like \(7\) will have a derivative of zero.

Step 2: Differentiate Each Term Separately

To differentiate each term:- For \(y = \frac{2}{5}x^3\), apply the power rule: \(\frac{d}{dx}(x^n) = nx^{n-1}\). Therefore, its derivative is \(\frac{2}{5} \times 3x^{2} = \frac{6}{5}x^2\).- Rewrite \(-\frac{4}{x^3}\) as \(-4x^{-3}\) and differentiate using the power rule: \(-3 \cdot -4x^{-4} = 12x^{-4}\).- For \(4\sqrt{x^5}\), rewrite as \(4x^{5/2}\) and differentiate: \(\frac{5}{2} \cdot 4x^{3/2} = 10x^{3/2}\).

Step 3: Combine the Differentiated Terms

Add the results of the derivatives together: \(\frac{6}{5}x^2 + 12x^{-4} + 10x^{3/2}\). This expression is the derivative of the original function.

Step 4: Simplify the Expression if Possible

In this case, the expression \(\frac{6}{5}x^2 + 12x^{-4} + 10x^{3/2}\) is already simplified as there are no like terms to combine.

Key Concepts

Differential Calculus

Differential calculus is all about understanding how functions change. At its core, it involves finding rates of change and slopes of curves. This is crucial in mathematics because it helps us understand everything from simple motions to complex systems. When we talk about the differential coefficient, we mean the derivative of a function—essentially measuring how the function value changes with respect to changes in the input. Knowing how to differentiate a function allows us to predict and analyze changes, an essential skill in fields like physics, engineering, and economics.

Power Rule

The power rule is a fundamental technique in differential calculus. It offers a straightforward way to differentiate polynomial functions, which are functions made up of terms like \(x^n\), where \(n\) is any real number. Here's how it works:

• Find the exponent of the term (the \(n\) in \(x^n\)).
• Multiply the term by this exponent.
• Subtract one from the exponent.

So for a function \(f(x) = x^n\), the derivative \(f'(x)\) is \(nx^{n-1}\). This formula makes it quick and easy to handle terms in a polynomial.

Derivative of Polynomial Functions

To differentiate polynomial functions, we apply the power rule to each term independently. Polynomials are simply sums of powers of \(x\) with coefficients—such as \(ax^n\), where \(a\) is another number. In our specific problem, the function is made up of several terms: \(\frac{2}{5}x^3\), \(-\frac{4}{x^3}\), and \(4\sqrt{x^5}\). Each term requires us to:

• Apply the power rule separately.
• Take into account constant multiples, like \(\frac{2}{5}\) or \(-4\).
• Simplify if needed after differentiation.

By understanding and applying the power rule, we can find derivatives of any polynomial expression efficiently.

Algebraic Manipulation

Algebraic manipulation involves rewriting terms to make differentiation easier. Sometimes, expressions might need to be rewritten to apply the power rule effectively.For instance, when given \(-\frac{4}{x^3}\), it can be rewritten as \(-4x^{-3}\) so the power rule can be applied directly. Similarly, transforming a radical expression like \(4\sqrt{x^5}\) into \(4x^{5/2}\) makes differentiation with the power rule straightforward. Algebraic manipulation is a helpful step in differential calculus as it simplifies the differentiation process, allowing students to tackle more complex expressions with confidence.