Question

Using the Product Rule In Exercises 1-6, use the Product Rule to find the derivative of the function. $$ g(x)=\sqrt{x} \sin x $$

Short answer

The derivative of the function g(x) is \(g'(x) = \frac{\sin x}{2\sqrt{x}} + \sqrt{x}\cos x\)

Step-by-step solution

Identify the functions u and v

\(\sqrt{x}\) and \(\sin x\) are the two functions that make up the function g. Here, let \(u=\sqrt{x}\) and \(v=\sin x\).

Differentiate the functions u and v

We need to find the derivatives of both functions. The derivative of \(u=\sqrt{x}\) is \(u'=\frac{1}{2\sqrt{x}}\) and the derivative of \(v=\sin x\) is \(v'=\cos x\).

Apply the Product Rule

Now we apply the Product Rule to compute the derivative of g: \(g'(x) = u'v + uv' = (\frac{1}{2\sqrt{x}}) * (\sin x) + (\sqrt{x}) * (\cos x) = \frac{\sin x}{2\sqrt{x}} + \sqrt{x}\cos x\).

Key concepts

Derivative of a Function

The derivative of a function at a point is a fundamental concept in calculus, providing the rate at which the function value changes as its input changes. To visualize this, imagine the function as representing the position of a moving object over time; the derivative then shows the velocity, indicating how fast the position is changing at any given moment.

To compute the derivative, we speak of the limit of the average rate of change as the interval over which the change is measured becomes infinitesimally small. The notation for the derivative of a function, such as \(f\), with respect to a variable, such as \(x\), is denoted \(f'(x)\), \(df/dx\), or \(\frac{d}{dx}f(x)\).

For example, if \(f(x) = x^2\), the derivative \(f'(x)\) is computed as \(2x\), showing that the rate of change of the square of \(x\) is proportional to \(x\) itself. Derivatives can also be interpreted as the slope of the tangent line to the curve of the function at any point, providing a geometric understanding of the rate of change.

Chain Rule

The Chain Rule is a powerful tool in calculus used to differentiate composite functions, where one function is applied within another. In essence, it connects the rates of change of the functions into a formula that gives the rate of change of the composite function.

For functions \(f\) and \(g\), where \(f\) is a function of \(g\), and \(g\) is a function of \(x\), the Chain Rule states that the derivative of the composite function \(f(g(x))\) with respect to \(x\) is \(f'(g(x))\) multiplied by \(g'(x)\). In notation, it is expressed as \(\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)\).

An analogy would be gears in a mechanism: if one gear turns another, the speed of the second gear depends on the speed of the first and the connection between them. Similarly, when applying the Chain Rule in calculus, the rate at which the outer function changes is multiplied by the rate at which the inner function changes, giving the overall rate of change of the composite function.

Trigonometric Functions Differentiation

Differentiating trigonometric functions is a recurring task in calculus, and understanding these derivatives is key to solving many problems involving periodic phenomena, such as waves. Each trigonometric function has its unique derivative:

• The derivative of \(sin(x)\) is \(cos(x)\).
• The derivative of \(cos(x)\) is \(-sin(x)\).
• The derivative of \(tan(x)\) is \(sec^2(x)\).

Just like with polynomial functions, these derivatives can be utilized to find the rate of change, or slope, for trigonometric functions at any point. For example, the derivative informs us that as \(x\) increases, the rate of increase of \(sin(x)\) is given by the value of \(cos(x)\) at that point.

Understanding these derivatives is crucial when dealing with more complex functions involving trigonometry, especially when combined with other calculus rules like the Product Rule or Chain Rule. For instance, to differentiate \(g(x) = sin(x)\cdot e^x\), you would need both the derivative of \(sin(x)\) and the application of the Product Rule to obtain the correct derivative of the function.