Question

Compute the derivative of the given function. $$g(t)=\sqrt{t} \sin t$$

Short answer

The derivative is \( g'(t) = \frac{1}{2 \sqrt{t}} \sin t + \sqrt{t} \cos t \).

Step-by-step solution

Identify the Function Type

The given function is a product of two functions: \( g(t) = \sqrt{t} \cdot \sin t \). To find the derivative, we need to use the product rule of differentiation.

Recall the Product Rule

The product rule states that if you have a function \( h(t) = u(t) \cdot v(t) \), then the derivative \( h'(t) = u'(t)v(t) + u(t)v'(t) \). Hence, we will apply this to \( g(t) = \sqrt{t} \cdot \sin t \).

Differentiate \( u(t) = \sqrt{t} \)

To differentiate \( \sqrt{t} \) with respect to \( t \), rewrite it as \( t^{1/2} \). Using the power rule, the derivative is \( u'(t) = \frac{1}{2} t^{-1/2} \).

Differentiate \( v(t) = \sin t \)

The derivative of \( \sin t \) with respect to \( t \) is \( v'(t) = \cos t \).

Apply the Product Rule

Substitute \( u(t) = \sqrt{t} \), \( u'(t) = \frac{1}{2} t^{-1/2} \), \( v(t) = \sin t \), and \( v'(t) = \cos t \) into the product rule formula.

Calculate \( g'(t) \) using Substituted Values

Applying the product rule: \[g'(t) = \left( \frac{1}{2} t^{-1/2} \right) \sin t + \sqrt{t} \cos t.\]Simplify the expression if needed: \[g'(t) = \frac{1}{2} t^{-1/2} \sin t + t^{1/2} \cos t.\]

Simplify the Derivative (If Needed)

The expression \( g'(t) = \frac{1}{2} \cdot t^{-1/2} \cdot \sin t + t^{1/2} \cdot \cos t \) is the simplified form of the derivative. Express \( t^{-1/2} \) and \( t^{1/2} \) as square roots if desired:\[ g'(t) = \frac{1}{2 \sqrt{t}} \sin t + \sqrt{t} \cos t \].

Key concepts

Product Rule

When dealing with derivatives of products of functions, the product rule is a crucial tool. Imagine two functions, let's say \( u(t) \) and \( v(t) \), and we're multiplying them to get a new function \( h(t) = u(t) \cdot v(t) \). The product rule helps us find the derivative \( h'(t) \) of this product.

This rule states that the derivative of the product \( h(t) \) can be found by differentiating each function separately and then applying this formula:

• Differentiate \( u(t) \) to get \( u'(t) \)
• Differentiate \( v(t) \) to get \( v'(t) \)
• Then, \( h'(t) \) is \( u'(t)v(t) + u(t)v'(t) \)

This rule allows us to handle complex expressions efficiently, providing a way to differentiate without needing to multiply out the expression first. It simplifies the process greatly by breaking it into manageable steps.

In our specific example, the function \( g(t) = \sqrt{t} \cdot \sin t \) was handled by identifying \( u(t) = \sqrt{t} \) and \( v(t) = \sin t \), differentiating them individually, and then applying their derivatives back in.

Differentiation

Differentiation is the process of finding the derivative of a function. A derivative represents the rate of change of a function concerning one of its variables. In simpler terms, it's like measuring how a given function "moves" or changes as the input changes just a little.

When we differentiate, we look for patterns or rules that tell us how to operate on functions to find this derivative. Calculus offers several methods, such as the product rule, quotient rule, and chain rule, for different situations.

In our given problem \( g(t)=\sqrt{t}\sin t \), finding the derivative \( g'(t) \) involves knowing some basic differentiation rules that fit the components of the function. We differentiate each part based on known rules and then combine them as necessary.

Power Rule

The power rule is one of the most fundamental rules in differentiation. It's especially handy because you often find yourself working with powers of variables. It says that for a function \( f(x) = x^n \), the derivative \( f'(x) \) is \( nx^{n-1} \).

For example, if you have \( x^3 \), the derivative would be \( 3x^{2} \). When you apply this rule, you're essentially subtracting one from the exponent and multiplying by the old exponent.

In our exercise, we use the power rule to differentiate \( \sqrt{t} \), which is essentially \( t^{1/2} \). Applying the power rule here, you get \( \frac{1}{2}t^{-1/2} \). This step is critical because it breaks down a potentially confusing expression into something more easily handled by following a straightforward rule.