91Ó°ÊÓ

In calculus, the Product Rule is used to find the derivative of the product of two functions. When you have two functions multiplied together, simply differentiating each and multiplying isn't accurate. Instead, the Product Rule comes to the rescue.Here's how it works:

• Let the two functions be denoted as \( u \) and \( v \).
• The derivative of their product \( uv \) with respect to \( x \) is given by: \( \frac{d}{dx}[uv] = u'v + uv' \).

Where \( u' \) and \( v' \) represent the derivatives of \( u \) and \( v \), respectively.
In the problem we have, the left side, \( x \cos(2x + 3y) \), uses the Product Rule because it's the product of two functions: \( x \) and \( \cos(2x + 3y) \).
This same concept applies to the right side, \( y \sin x \), where \( y \) and \( \sin x \) are multiplied.

The Chain Rule is a powerful technique in calculus for finding the derivative of composite functions. When a function is applied within another function, the Chain Rule is necessary to correctly calculate the derivative.Here's how it is applied:

• If you have a function \( y = f(g(x)) \), the derivative of \( y \) with respect to \( x \) is \( f'(g(x)) \cdot g'(x) \).

In the given problem, we used the Chain Rule while differentiating \( \cos(2x + 3y) \) and \( y \sin x \):

• For \( \cos(2x + 3y) \), view \( 2x + 3y \) as an "inside" function. The derivative is \( -\sin(2x + 3y) \cdot (2 + 3 \frac{dy}{dx}) \).
• This part \( (2 + 3 \frac{dy}{dx}) \) comes from differentiating \( 2x + 3y \) since \( y \) is also a function of \( x \).

The case of \( y \sin x \) is slightly different but still involves recognizing the dependence of \( y \) on \( x \).

Finding derivatives is a fundamental aspect of calculus. Derivatives measure how a function changes as its input changes. Implicit Differentiation is a method used when finding derivatives in equations where the variables are intertwined, unlike straightforward explicit functions.Here's the step-by-step approach for implicit differentiation:

• Differentiate each term of the equation with respect to \( x \).
• Treat \( y \) as a function of \( x \), which means when you differentiate \( y \), you multiply by \( \frac{dy}{dx} \).

In the exercise, after differentiating the whole equation, we make use of algebra to solve for \( \frac{dy}{dx} \).
This involves gathering all the terms containing \( \frac{dy}{dx} \) on one side to isolate it, which results in the derivative of \( y \) in terms of \( x \).
The final result is achieved by factoring and isolating \( \frac{dy}{dx} \), allowing us to solve for it systematically and precisely.