91Ó°ÊÓ

When you have two differentiable functions, like \( u(x) \) and \( v(x) \), and you want to differentiate their product, the Product Rule is your friend. The Product Rule states that the derivative of the product \( uv \) is not just the product of their derivatives. Instead, the formula is:

• \[ \frac{d}{dx}(uv) = u'v + uv' \]

Here, \( u' \) and \( v' \) denote the derivatives of \( u \) and \( v \) respectively. This way, even though both \( u \) and \( v \) are changing, we account for their mutual effect on the derivative by applying the rule.
This rule is particularly useful when you have a product of functions where each function might change its behavior over time.
In the exercise example, substituting \( u'(1) = 0 \), \( v(1) = 5 \), \( u(1) = 2 \), and \( v'(1) = -1 \) into the Product Rule, we find that the derivative at \( x = 1 \) is \( -2 \). This calculation shows how changes in both functions combine to influence the overall change of their product.

While the Product Rule helps with products, the Quotient Rule handles division. When differentiating the quotient \( \frac{u}{v} \) of two functions \( u(x) \) and \( v(x) \), the rule is slightly more complex:

• \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]

This formula comes in handy because it not only calculates the change in the numerator but also considers how changes in the denominator alter the whole expression.
In our example, applying the given derivative values, the derivative of \( \frac{u}{v} \) at \( x = 1 \) is calculated as \( \frac{2}{25} \). This result verifies how the dynamic nature of both \( u \) and \( v \) affects the rate of change of their ratio.

Sometimes, expressions involve combinations of functions that are straightforwardly algebraic. These are known as linear combinations. A linear combination of functions \( u \) and \( v \) looks like \( au + bv \) where \( a \) and \( b \) are constants.

• The derivative of \( au + bv \) is simply \( au' + bv' \).

This rule is derived from the linearity of differentiation from basic calculus principles. It’s a simplified application because differentiation affects the functions regardless of scalar multiplication.
In the problem, \( 7v - 2u \) falls into this category. By applying the linearity property, the derivative was found to be \( 7v' - 2u' \). With our values, this simplifies to \( -7 \), illustrating the simplicity of handling constants and functions together in derivatives.

In calculus, understanding differentiability is crucial. A function is differentiable at a point if it has a derivative there, meaning it smoothly changes around that point without any abrupt turns or breaks.

• Both \( u(x) \) and \( v(x) \) being differentiable indicates they can be operated upon using differentiation rules like Product and Quotient.

Differentiability ensures that functions comply with all the rules seamlessly, providing reliable computations of their change rates.
In the exercise, this property allowed us to apply various differentiation rules such as the Product and Quotient rules without issue. Both \( u \) and \( v \) having derivatives at \( x = 1 \) means they are smooth and predictable functions at that point, making it possible to determine derivatives of their combinations confidently and accurately.