91Ó°ÊÓ

When you come across a function that involves the multiplication of two separate functions, it's crucial to use the Product Rule to find the derivative effectively.
The Product Rule states that if you have two functions, say \( u \) and \( v \), the derivative of their product \( u \cdot v \) is given by:

• \( (u \cdot v)' = u' \cdot v + u \cdot v' \)

Here, \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.
To make this concept tangible, let's explore it with an example from the original exercise. We have \( y = x \arctan(2x) \), where \( u = x \) and \( v = \arctan(2x) \).

• The derivative of \( u \), which is \( u' = 1 \).
• The derivative of \( v \), which is \( v' = \frac{2}{1+4x^2} \).

So, applying the Product Rule gives us:
\( y' = 1 \cdot \arctan(2x) + x \cdot \frac{2}{1+4x^2} \)
This approach allows you to handle any product of functions by breaking down each part systematically.

Logarithmic differentiation is an efficient method for differentiating functions that involve a logarithm, especially when the function is complex.
The standard rule for differentiating a natural logarithm \( \ln(u) \) is to find \( \frac{u'}{u} \).
To solidify this concept, consider the logarithmic part of the function from our exercise: \(-\frac{1}{4} \ln(1 + 4x^2)\).
Here, \( u = 1 + 4x^2 \), therefore \( u' = 8x \).
Applying the rule we get:

• \( \frac{-1}{4} \cdot \frac{8x}{1+4x^2} \)

Simplifying, you have \( -\frac{2x}{1+4x^2} \).
Logarithmic differentiation is quite powerful as it simplifies the differentiation of products, quotients, or powers by transforming them into sums.

Trigonometric functions, such as \( \arctan \), play an important role in calculus, particularly when dealing with derivatives.
The derivative of the inverse trigonometric function \( \arctan(x) \) is \( \frac{1}{1+x^2} \).
In our exercise, the trigonometric component is \( \arctan(2x) \).
To differentiate \( \arctan(2x) \), we first apply the chain rule, since it's a function of a function:

• Set \( g(x) = 2x \), then \( g'(x) = 2 \).
• The derivative of \( \arctan(g(x)) \) is \( \frac{g'(x)}{1+(g(x))^2} = \frac{2}{1 + (2x)^2} \).

This gives us \( v' = \frac{2}{1+4x^2} \), just as found in the exercise.
Understanding how to differentiate trigonometric functions unlocks answers to many calculus problems involving angles and rates.