Question

Working with Numerical Values Suppose that functions \(f\) and \(g\) and their derivatives have the following values at \(x=2\) and \(x=3\) $$\begin{array}{c|cccc}{x} & {f(x)} & {g(x)} & {f^{\prime}(x)} & {g^{\prime}(x)} \\ \hline 2 & {8} & {2} & {1 / 3} & {-3} \\ {3} & {3} & {-4} & {2 \pi} & {5}\end{array}$$ Evaluate the derivatives with respect to \(x\) of the following combinations at the given value of \(x .\) (a) 2\(f(x)\) at \(x=2\) (b) \(f(x)+g(x)\) at \(x=3\) (c) \(f(x) \cdot g(x)\) at \(x=3\) (d) \(f(x) / g(x)\) at \(x=2\) (e) \(f(g(x))\) at \(x=2\) (f) \(\sqrt{f(x)}\) at \(x=2\) (g) 1\(/ g^{2}(x)\) at \(x=3\) (h) \(\sqrt{f^{2}(x)+g^{2}(x)} \quad\) at \(\quad x=2\)

Short answer

The derivatives are: (a) 2/3, (b) \(2\pi + 5\), (c) \(15 - 8\pi\), (d) \(38/3\), (e) -1, (f) 1/6\(\sqrt{8}\), (g) -5/16, (h) -11/3\(\sqrt{68}\).

Step-by-step solution

Find the derivative of 2\(f(x)\) at \(x=2\)

The derivative of a function multiplied by a constant, 2 in this case, is the constant times the derivative of the function. So the derivative of 2\(f(x)\) at \(x=2\) is 2 times \(f'(2)\), which equals 2 * \(1/3 = 2/3\)

Find the derivative of \(f(x)+g(x)\) at \(x=3\)

The derivative of the sum of two functions is the sum of their derivatives. Therefore, the derivative of \(f(x)+g(x)\) at \(x=3\) is \(f'(3) + g'(3) = 2\pi + 5\)

Find the derivative of \(f(x) \cdot g(x)\) at \(x=3\)

For the formula of the derivative of a product of two functions, we use the product rule: \((f(x) \cdot g(x))' = f(x) \cdot g'(x) + g(x) \cdot f'(x)\). Applying the rule at \(x=3\) gives \(f(3) \cdot g'(3) + g(3) \cdot f'(3) = 3 * 5 - 4 * 2\pi\)

Find the derivative of \(f(x) / g(x)\) at \(x=2\)

For the formula of the derivative of a quotient of two functions, use the quotient rule: \( (f(x)/g(x))' = (g(x)\cdot f'(x) - f(x)\cdot g'(x)) / (g(x))^2 \). Applying the rule at \(x=2\) gives \((2 \cdot 1/3 - 8 \cdot -3) / 2^2 = 2/3 + 12 = 38/3\)

Find the derivative of \(f(g(x))\) at \(x=2\)

The Chain Rule works in cases where we have a composition of two functions, as in \(f(g(x))\). We need \(f'(g(x))\) and \(g'(x)\). This derivative at \(x=2\) is \(f'(2) \cdot g'(2) = 1/3 * -3 = -1\)

Find the derivative of \(\sqrt{f(x)}\) at \(x=2\)

The derivative of the square root of a function is \(1/2 * f'(x) / \sqrt{f(x)}\). Finding this derivative at \(x=2\) gives \(1/2*1/3/\sqrt{8} = 1/6\sqrt{8}\)

Find the derivative of 1\(/ g^{2}(x)\) at \(x=3\)

The derivative is given by \(-f'(x)/(g(x))^2\). So we need to substitute with the given values, \(f'(3) = -5\), \(g(3) = -4\). Hence the derivative at \(x=3\) is \(-5/(-4)^2 = -5/16\)

Find the derivative of \(\sqrt{f^{2}(x)+g^{2}(x)} \) at \(x=2\)

The derivative formula is \( (f(x)f'(x) + g(x)g'(x)) / \sqrt{f^2(x) + g^2(x)}\). Substituting the given values gives \( (8*1/3 + 2*-3) / \sqrt{8^2 + 2^2} = -11/3\sqrt{68}\)

Key concepts

Product Rule

When calculating the derivative of a product of two functions, we apply what is known as the product rule. This rule is particularly useful when two distinct functions are multiplied together. It's expressed mathematically as \( (f(x) \cdot g(x))' = f(x) \cdot g'(x) + g(x) \cdot f'(x) \).

Let's take, for instance, our exercise example \( f(x)g(x) \) at \( x=3 \). We use the values \( f(3) \) and \( g(3) \) along with their derivatives \( f'(3) \) and \( g'(3) \) to find the derivative of the product. The computation becomes a simple plug-and-play, resulting in \( f(3) \cdot g'(3) + g(3) \cdot f'(3) \), which is \(-12 + 15\) after substituting the appropriate values.

Quotient Rule

In differentiation, when we have a function that is the quotient of two other functions, we use the quotient rule to find the derivative. This rule can be represented as \( (\frac{f(x)}{g(x)})' = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{\left(g(x)\right)^2} \).

As an example, to find the derivative of \( \frac{f(x)}{g(x)} \) when \( x=2 \), we plug the values of \( f(2) \) and \( g(2) \) and their derivatives into the formula. A note of caution here: always make sure your denominator is squared properly to avoid any calculation errors. In our exercise, the final calculation looks like \( (2 \cdot \frac{1}{3} + 8 \cdot 3) \big/ 4 \) after plugging in the values from the table.

Chain Rule

The chain rule is a fundamental differentiation technique used when dealing with composite functions, where one function is nested inside another. The formal notation is \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).

In practice, we find the derivative of the outer function as if the inner function were a variable, and then multiply it by the derivative of the inner function. For instance, in our exercise we were given \( f(g(x)) \) at \( x=2 \). Thus, we calculated \( f'(g(2)) \cdot g'(2) \) by using the derivative values provided, leading us to a solution of \( -1 \) once we made our substitutions.

Derivative of Sums

When we encounter the sum of multiple functions and need to find its derivative, the rule is straightforward and intuitive: simply take the derivative of each function individually and then sum the results. This rule comes from the linearity of differentiation, which can be summed up by \( (f(x) + g(x))' = f'(x) + g'(x) \).

For example, for the sum \( f(x) + g(x) \) at \( x=3 \) in our exercise, the derivative was \( f'(3) + g'(3) \) which adds up to \( 2\pi + 5 \), demonstrating the concept's practical application and simplicity.

Differentiation Techniques

Differentiation techniques encompass a series of rules and formulas that enable us to compute the derivatives of various types of functions. Beyond the product, quotient, and chain rules, we have other techniques such as the power rule, derivatives of trigonometric functions, implicit differentiation, and higher-order derivatives.

These techniques are tools in our mathematical toolbox, allowing us to tackle more complex functions. For example, in our exercise, different techniques were used to calculate derivatives for various functions given specific values for \( x \) with the goal of making complex calculations manageable. Each technique is suited to a particular function type, ensuring we can find derivatives efficiently and accurately.