Problem 44 - Calculus of a Single Variable

Question

Finding a Derivative In Exercises \(33-54,\) find the derivative. $$ g(t)=e^{-3 / t^{2}} $$

Step-by-Step Solution

Answer

The derivative of \(g(t) = e^{-3/t^2}\) is \(g'(t) = 6e^{-3 / t^{2}} / t^{3}\).

Step 1: Understand the function

The function given is \(g(t) = e^{-3/t^2}\). Here we have the number \(e\) raised to the power of \(-3/t^2\).

Step 2: Apply the chain rule

When differentiating a function that is composed of two or more functions, it is necessary to apply the chain rule. The chain rule in differentiation states if a variable \(x\) is a function of \(y\), and \(y\) in turn is a function of \(z\), then the derivative of \(x\) with respect to \(z\) can be calculated as such: \(x' = x'_y * y'_z\). In this case, we can represent our function in the form of a composition of two functions: \(y = -3/t^2\) and \(x = e^y\). Hence \((e^y)'_t = (e^y)'_y * y'_t\). Let's find the derivative of \(e^y\) with respect to \(y\), which will denote as \((e^y)'_y\), and the derivative of \(y\) (\(-3/t^2\)) with respect to \(t\), which will denote as \(y'_t\).

Step 3: Differentiate e^y

The derivative of \(e^y\) with respect to \(y\) is simply \(e^y\), so \((e^y)'_y = e^y\).

Step 4: Differentiate y

Now differentiate \(-3/t^2\) with respect to \(t\). This can be written as \(-3 * t^{-2}\). Using the power rule in differentiation, we get \(y'_t = 6 / t^3\).

Step 5: Apply the results into the chain rule

Now substitute \((e^y)'_y\) and \(y'_t\) into the chain rule equation. We get \((e^y)'_t = e^y * (6 / t^3) = (6 / t^3) * e^{-3/t^2}\).

Step 6: Simplify

Simplifying the expression, we get \(g'(t) = 6e^{-3 / t^{2}} / t^{3}\).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, especially when dealing with composite functions. It helps us find the derivative of a function that is composed of two or more functions. Imagine you have a function inside another function—like a layered cake. The chain rule essentially tells us how to "de-layer" these functions to differentiate them efficiently.
When applying the chain rule, we need to differentiate the outer function first, and then multiply it by the derivative of the inner function. For example, if we have a function like \(g(x) = f(u(x))\), then the derivative \(g'(x)\) is found by multiplying \(f'(u(x))\) by \(u'(x)\).

• Differentiate the outer function (leaving the inner function intact).
• Multiply by the derivative of the inner function.

This method is essential for functions like exponential functions, trigonometric functions, and logarithmic functions, when they are not in their simplest form.

Exponential Functions

Exponential functions, especially those involving the natural base \(e\), are important in calculus due to their unique properties. These functions have the form \(f(x) = e^{u(x)}\), where \(u(x)\) can be any function of \(x\).
The fascinating aspect of the natural exponential function \(e^x\) is that its rate of change, or derivative, is the same as the function itself, \(\frac{d}{dx}e^x = e^x\).
When dealing with more complex exponential functions like \(e^{-3/t^2}\):

• Recognize the exponential function structure.
• Identify the inner function, here it's \(-3/t^2\).

The derivative of such a function follows the chain rule, as mentioned earlier. This allows us to handle exponential growth and decay processes in fields like biology, physics, and economics.

Differentiation Techniques

Differentiation techniques are the methods used to find the derivative of a function. In calculus, derivatives are key to understanding how a function changes at any point.
In the problem, two main techniques are used:

• Chain Rule: As discussed, this is crucial when differentiating composite functions.
• Power Rule: This rule simplifies finding derivatives of functions like \(x^n\). For example, for \(-3/t^2\), we rewrite it as \(-3t^{-2}\) and apply the power rule to get \(6t^{-3}\). This is equivalent to \(6/t^3\).

Learning to combine these techniques is vital in calculus, as real-world problems often involve complex functions. By mastering these, you can tackle problems in physics, engineering, and beyond.