Question

Compute the derivative of the given function. $$g(t)=\cos \left(\frac{1}{t}\right) e^{5 t^{2}}$$

Short answer

The derivative is \( \sin\left(\frac{1}{t}\right)\frac{e^{5t^2}}{t^2} + 10t\cos\left(\frac{1}{t}\right)e^{5t^2}\).

Step-by-step solution

Identify the function type

The given function is a product of two functions: \( u(t) = \cos \left( \frac{1}{t} \right) \) and \( v(t) = e^{5t^2} \). We need to use the product rule to find the derivative of \( g(t) = u(t) \cdot v(t) \).

Apply the Product Rule

The product rule states that for two functions \( u(t) \) and \( v(t) \), the derivative \( g'(t) = u'(t) \cdot v(t) + u(t) \cdot v'(t) \). We will find \( u'(t) \) and \( v'(t) \) next.

Differentiate \( u(t) = \cos \left( \frac{1}{t} \right) \)

To differentiate \( u(t) = \cos \left( \frac{1}{t} \right) \), use the chain rule. Set \( z = \frac{1}{t} \), then \( u(t) = \cos(z) \). The derivative is \( u'(t) = -\sin(z) \cdot \frac{d}{dt}(z) \). With \( \frac{d}{dt}(z) = -\frac{1}{t^2} \), we find \( u'(t) = \sin \left( \frac{1}{t} \right) \cdot \frac{1}{t^2} \).

Differentiate \( v(t) = e^{5t^2} \)

For \( v(t) = e^{5t^2} \), apply the chain rule. The exponential derivative is \( v'(t) = e^{5t^2} \cdot \frac{d}{dt}(5t^2) = e^{5t^2} \cdot 10t \).

Substitute derivatives into the product rule formula

Using the product rule, \( g'(t) = u'(t) \cdot v(t) + u(t) \cdot v'(t) \), substitute \( u'(t) = \sin \left( \frac{1}{t} \right) \cdot \frac{1}{t^2} \), \( u(t) = \cos \left( \frac{1}{t} \right) \), \( v(t) = e^{5t^2} \), and \( v'(t) = 10t \cdot e^{5t^2} \).

Simplify the expression

Combine the expressions: \( g'(t) = \left( \sin \left( \frac{1}{t} \right) \cdot \frac{1}{t^2} \right) e^{5t^2} + \cos \left( \frac{1}{t} \right) \cdot \left(10t \cdot e^{5t^2} \right) \). This simplifies to \( g'(t) = \sin \left( \frac{1}{t} \right) \cdot \frac{e^{5t^2}}{t^2} + 10t \cos \left( \frac{1}{t} \right) \cdot e^{5t^2} \).

Key concepts

Product Rule

The product rule is essential when dealing with the derivative of a product of two differentiable functions. In this exercise, we have the function \( g(t) = u(t) \cdot v(t) \) where \( u(t) = \cos \left( \frac{1}{t} \right) \) and \( v(t) = e^{5t^2} \).
To apply the product rule, remember that the derivative \( g'(t) \) is calculated as:

• \( g'(t) = u'(t) \cdot v(t) + u(t) \cdot v'(t) \)

This means, you need to find the derivatives of both \( u(t) \) and \( v(t) \) separately and then utilize this formula.
As illustrated, once you have the derivatives, you can substitute them in and simplify the expression. This technique allows you to handle functions multiplied together efficiently.

Chain Rule

The chain rule is your best friend when differentiating composite functions. In this case, both \( u(t) = \cos \left( \frac{1}{t} \right) \) and \( v(t) = e^{5t^2} \) involve composite functions, so you need the chain rule.
For \( u(t) \), by setting \( z = \frac{1}{t} \), you find that \( u(t) = \cos(z) \). The chain rule formula \( u'(t) = -\sin(z) \cdot \frac{d}{dt}(z) \) helps you substitute that \( \frac{d}{dt}(z) = -\frac{1}{t^2} \), leading to:

• \( u'(t) = \sin \left( \frac{1}{t} \right) \cdot \frac{1}{t^2} \)

For \( v(t) = e^{5t^2} \), apply the chain rule as follows:

• \( v'(t) = e^{5t^2} \cdot \frac{d}{dt}(5t^2), \, with \, \frac{d}{dt}(5t^2) = 10t \)
• \( v'(t) = e^{5t^2} \cdot 10t \)

Thus, the chain rule simplifies dealing with functions inside functions.

Trigonometric Functions

Trigonometric functions such as cosine and sine are prevalent in calculus for modeling periodic behaviors. Here, the trigonometric function \( \cos \left( \frac{1}{t} \right) \), which is part of \( u(t) \), shows how sine and cosine derivatives play:

• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• For \( u(t) \), transform \( \cos \left( \frac{1}{t} \right) \) using \( z = \frac{1}{t} \) and differentiate as explained.

These derivatives are especially useful in solving engineering and physics problems that involve oscillations or wave-like patterns.

Exponential Functions

Exponential functions are fundamental due to their growth properties. In this exercise, the function \( v(t) = e^{5t^2} \) is an exponential function with a compound argument \( 5t^2 \).
The basic derivative rule for exponential functions \( \frac{d}{dx}(e^{f(x)}) = e^{f(x)} \cdot f'(x) \) allows us to calculate:

• The derivative is \( v'(t) = e^{5t^2} \cdot 10t \), where the term \( 10t \) comes from differentiating \( 5t^2 \).

Exponential functions like these frequently describe natural phenomena such as population growth or radioactive decay, hence their importance.