Problem 4 - Single Variable Calculus

Question

Find the derivative of the function. $$ y=\sqrt{x} \sin x $$

Step-by-Step Solution

Answer

The derivative of the function \(y = \sqrt{x}\sin x\) is: $$ y' = \frac{1}{2}\sin x\cdot x^{-\frac{1}{2}}+\sqrt{x}\cos x $$

Step 1: Identify the Functions

Defined the functions \(u\) and \(v\), where $$ u(x)=\sqrt{x} \\ v(x)=\sin x $$

Step 2: Compute the Derivatives of u and v

For the function $$u(x)=\sqrt{x}$$, compute the derivative using the power rule: $$ u'(x)=\frac{d}{dx}\left( x^{\frac{1}{2}} \right)=\frac{1}{2}x^{-\frac{1}{2}} $$ For the function $$v(x)=\sin x$$, compute the derivative using the identity \( \frac{d(\sin x)}{dx}=\cos x\): $$ v'(x)=\cos x $$

Step 3: Apply the Product Rule

Using the product rule and previously calculated derivatives, compute the derivative of the given function: $$ y'=\left(\frac{1}{2}x^{-\frac{1}{2}}\right)(\sin x)+(\sqrt{x})(\cos x) $$

Step 4: Simplify the Solution

Rearrange the terms to achieve the final derivative: $$ y'=\frac{1}{2}\sin x\cdot x^{-\frac{1}{2}}+\sqrt{x}\cos x $$ The derivative of the function \(y = \sqrt{x}\sin x\) is: $$ y' = \frac{1}{2}\sin x\cdot x^{-\frac{1}{2}}+\sqrt{x}\cos x $$

Key Concepts

Product Rule

When faced with a function that is the product of two other functions, like in the exercise above where we have a function of the form \(y = u(x)v(x)\), we apply the product rule to find the derivative. The product rule is a fundamental technique in calculus for differentiating products of two functions. It states that the derivative of the product of two functions is given by:

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

This means you take the derivative of the first function (\(u'\)) and multiply it by the second function (\(v\)), and then you add the product of the first function (\(u\)) with the derivative of the second function (\(v'\)). In the exercise, to find \(y'\), you calculate the derivatives of \(u(x) = \root{x}\) and \(v(x) = \root{x}\), then apply the product rule as outlined in the solution.

Power Rule

The power rule is key for differentiating functions that are power expressions of the form \(x^n\), where \(n\) is any real number. According to this rule, the derivative of such a function is:\[\frac{d}{dx}(x^n) = nx^{n-1}\]

To apply the power rule to \(u(x) = \root{x}\), we rewrite \(\root{x}\) as \(x^{1/2}\). So, the derivative of \(u(x)\), which we denote as \(u'(x)\), is \(\frac{1}{2}x^{\frac{1}{2}-1} = \frac{1}{2}x^{-\frac{1}{2}}\), as demonstrated in the given solution. The power rule simplifies the process of differentiating polynomials significantly, making it a foundational tool in calculus.

Derivative of Sine

The sine function, which is a basic trigonometric function, also has a straightforward derivative. The derivative of the sine function is always the cosine function. In mathematical terms, for a function \(v(x) = \root{x}\), this relationship is given by:\[v'(x) = \frac{d}{dx}(\root{x}) = \root{x}\]

Utilizing this identity, the differentiation of \(\sin x\) in our exercise becomes a simple task of recognizing that the derivative is \(\cos x\). Trigonometric functions like sine and cosine are periodic, and their derivatives are well-defined, providing a predictable pattern that is useful in solving calculus problems.

Chain Rule

The chain rule is a valuable tool in calculus for finding the derivative of a composite function. In simpler terms, if you have a function that is the combination of two functions, the chain rule allows you to differentiate it efficiently. The formal expression for the chain rule is:\[\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)\]

Although it wasn't directly applied in this exercise, understanding the chain rule is essential, especially if either \(u(x)\) or \(v(x)\) were more complex functions involving compositions. The chain rule helps you 'peel back the layers' of the composite functions, differentiating from the outer function to the inner function sequentially.