Question

If \(f(t)=5 t+\frac{1}{\sqrt{t^{3}}}\) find \(f^{\prime}(t)\)

Short answer

The derivative is \( f'(t) = 5 - \frac{3}{2t^{5/2}} \).

Step-by-step solution

Differentiate the Linear Term

The function given is \( f(t) = 5t + \frac{1}{\sqrt{t^3}} \). The first term is \( 5t \). The derivative of a linear function \( at \) with respect to \( t \) is simply the constant \( a \). So, for \( 5t \), the derivative is \( 5 \).

Rewrite the Second Term for Differentiation

The second term is \( \frac{1}{\sqrt{t^3}} \). This can be rewritten as \( t^{-3/2} \) by using the rule that \( \sqrt{t^3} = t^{3/2} \) and then applying the reciprocal rule to get \( t^{-3/2} \).

Differentiate the Rewritten Second Term

Using the power rule for differentiation, \( \frac{d}{dt} [t^n] = n \cdot t^{n-1} \), we differentiate \( t^{-3/2} \). Applying the power rule, we get the derivative: \( -\frac{3}{2}t^{-3/2-1} = -\frac{3}{2}t^{-5/2} \).

Combine the Derivatives

Now we combine the derivatives from Step 1 and Step 3. The derivative of the function \( f(t) \) is \( f'(t) = 5 + \left(-\frac{3}{2}t^{-5/2}\right) \). This simplifies to \( f'(t) = 5 - \frac{3}{2t^{5/2}} \).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus. It is used to find the rate at which a function is changing at any given point. Imagine a car traveling along a road; differentiation helps us find out how fast the car is moving at any specific instant. In mathematical terms, it provides us with the derivative of a function. This process breaks down complex functions into simpler ones, enabling us to analyze their behavior more easily. For this exercise, remember the function involves both a linear and a non-linear term, and each part requires handling them separately during differentiation.

Power Rule

The power rule is one of the most straightforward techniques for differentiation. It allows us to find the derivative of polynomial expressions easily.
This rule states that if you have a function expressed as a power, like \(t^n\), its derivative is \(n \cdot t^{n-1}\). This means you bring down the exponent as a coefficient and subtract one from the original exponent.

• For example, the derivative of \(t^3\) is \(3t^2\).
• The derivative of \(t^{-4}\) is \(-4t^{-5}\).

In our problem, once we have rewritten the tricky fraction term using exponents, the power rule makes finding its derivative straightforward.

Derivative of Functions

The derivative of a function provides critical information about the function's behavior—such as its increasing or decreasing behavior at certain intervals. For the given function, \(f(t) = 5t + \frac{1}{\sqrt{t^3}}\), we have to find its derivative \(f'(t)\).
Here, each part of the function contributes to the total derivative:

• For \(5t\), the derivative is a simple constant due to the power of 1.
• For the complex term \(\frac{1}{\sqrt{t^3}}\), rewriting it helps simplify differentiation through the power rule.

The derived result of a function tells us how each small change in \(t\) affects \(f(t)\). This operation helps analyze real-world problems involving changes and rates.

Rewriting Expressions

Rewriting expressions in a differentiable form is a useful strategy in calculus because it simplifies the process of finding derivatives. In this exercise, the term \(\frac{1}{\sqrt{t^3}}\) was initially challenging to differentiate. By rewriting it as \(t^{-3/2}\), it becomes more manageable.

• Using rules of exponents, \(\sqrt{t^3}\) becomes \(t^{3/2}\).
• Applying the reciprocal, \(\frac{1}{t^{3/2}}\) can be rewritten as \(t^{-3/2}\).

This transformation makes it suitable for applying the power rule. Rewriting expressions is a crucial skill in calculus, greatly easing the process of computing derivatives for complex functions. Mastery of this skill lets you tackle a wide variety of mathematical challenges.