91Ó°ÊÓ

A derivative represents how a function changes as its input changes. In simple terms, it's like finding the slope of a curve at any point. For a function of a single variable, the first derivative tells us the rate at which the function value is changing. In the given exercise, we first found derivatives of x and y with respect to the parameter t. Let’s break it down:
We have two functions:

• \(x = t^2 e^t\)
• \(y = t \, \ln t\)

To compute the first derivatives, we use rules like the product rule:

• \(\frac{dx}{dt} = \frac{d}{dt}(t^2 e^t)\)
• \(\frac{dy}{dt} = \frac{d}{dt}(t \, \ln t)\)

This way we found \(\frac{dx}{dt} = e^t (t^2 + 2t)\) and \( \frac{dy}{dt} = \ln t + 1 \).
Now, the first derivative of y with respect to x can be found using the chain rule.

The second derivative provides the rate of change of the rate of change (essentially the acceleration) of a function. It can give us important information about the concavity and points of inflection of a function.
In the given problem, after finding \( \frac{dy}{dx} \), we needed to compute the second derivative \( \frac{d^2 y}{dx^2} \). To do this, we differentiate \( \frac{dy}{dx} \) with respect to t and then divide by \( \frac{dx}{dt} \) again using the quotient rule:

• Let’s say \(u = \ln t + 1\)
• and \(v = e^t (t^2 + 2t)\)

Using the quotient rule:
\[ \frac{d}{dt} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \]
We then compute \( u' \) and \( v' \):

• \( u' = \frac{1}{t} \)
• \( v' = e^t (t^2 + 4t + 2) \)

Combining these results through the quotient rule, we obtain the second derivative \( \frac{d^2 y}{dx^2} \).

The chain rule is a fundamental tool in calculus used to differentiate composite functions. It states that to differentiate a composite function, you first differentiate the outer function and then multiply it by the derivative of the inner function.
Mathematically, if \( y = f(g(t)) \), then the derivative \( \frac{dy}{dx} = f'(g(t))\cdot g'(t) \).
In our exercise, to find \( \frac{dy}{dx} \), we needed to relate \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \):

• \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \)

We already found \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):

• \( \frac{dx}{dt} = e^t (t^2 + 2t) \)
• \( \frac{dy}{dt} = \ln t + 1 \)

Thus, \( \frac{dy}{dx} = \frac{\ln t + 1}{ e^t (t^2 + 2t) } \).

The quotient rule is used for finding the derivative of a function that is the ratio of two differentiable functions. It is particularly useful when we have a division of two functions.
If you have functions \( f(t)\) and \( g(t) \) and you want to differentiate \( \frac{f(t)}{g(t)} \), the quotient rule states:
\[ \left( \frac{f}{g} \right)' = \frac{f' g - f g'}{g^2} \]
In the solution, we applied the quotient rule to find the second derivative:
Given \( u = \ln t + 1 \) and \( v = e^t (t^2 + 2t) \), the derivative is:
\( \frac{d}{dt} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \).
With \( u' = \frac{1}{t} \) and \( v' = e^t (t^2 + 4t + 2) \), we apply these to the quotient rule to get the second derivative.