Suppose that functions f and g and their derivatives with...

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

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

Key Concepts

Chain Rule

The Chain Rule is essential for differentiating composite functions, where one function is nested within another. It states that the derivative of the composite function f(g(x)) is f'(g(x)) multiplied by g'(x), thus linking the rates of change of the inner and outer functions.

Quotient Rule

This rule is applied when differentiating a quotient of two functions. It states that the derivative of f(x)/g(x) is (f'(x)·g(x) - f(x)·g'(x)) divided by [g(x)]². This formula accounts for the dependence of both the numerator and the denominator on x.

Power Rule

The Power Rule is a basic differentiation rule used when a function is raised to a power. When combined with the Chain Rule, it also applies to expressions where the function inside the power is not just the variable x. For example, in functions like sqrt(f(x)) or 1/g(x)^2, the rule is used in conjunction with the Chain Rule.

Sum Rule

The Sum Rule indicates that the derivative of a sum of functions is the sum of their individual derivatives. This rule makes it straightforward to differentiate expressions composed of multiple terms added or subtracted, such as f(x) + g(x).

Product Rule

The Product Rule is used when differentiating the product of two functions. It states that the derivative of f(x)·g(x) is given by f'(x)·g(x) + f(x)·g'(x), highlighting that both the derivatives of each function and the original functions themselves must be considered.

Constant Multiple Rule

This rule states that the derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function. It simplifies differentiating expressions like c·f(x) by allowing the constant to be factored out before taking the derivative.

Transcript

00:01 Okay, this is a big long problem.

00:03 So 3 .6, problem number 75a.

00:08 So they're going to give us this information about f and g and the derivatives, and they want us to find the derivatives at certain points.

00:16 So we're asked to differentiate 2f of x when x is equal to 2.

00:23 Okay.

00:24 So if i differentiate this, i've just got a constant in front of this.

00:28 So this is going to be 2 f prime of x.

00:32 So what happens when x is equal to 2? well, i would need to know what is f prime of 2.

00:39 F prime of 2 is 1 3rd.

00:41 So this is 2 times 1 3rd, which is 2 thirds.

00:47 Okay, that is part a.

00:50 If you look at part b, so in part b, they give us f of x plus g of x.

00:56 So f of x plus g of x and they want us to find the derivative of this evaluated when x is equal to three well adding these things together that's just f prime of x plus g prime of x so evaluate this at x equal to three that is f prime of three plus g prime of three which is equal to so so f prime of 3 is 2 pi and g prime of 3 is 5.

01:41 So 2 pi plus 5 is the answer to part b.

01:48 Part c, so in part c they give us f of x times g of x.

01:54 So f of x times g of x.

01:59 And we want to know the derivative of this evaluated when x is equal to 3.

02:08 So this is just simply the product rule.

02:11 So to differentiate this product, we're going to have f prime of x, g of x, plus g prime of x, f of x.

02:23 And we are asked to evaluate this when x is equal to three.

02:27 So that is going to be f prime of three times g of three plus g prime of three times f of three.

02:39 So f prime of 3, that is 2 pi, g of 3, so g of 3 is negative 4, g prime of 3, g prime of 3 is 5, and f of 3 is 3.

03:01 So this is going to be about 15 minus 8 pi.

03:06 So 15 minus 8 pi is the answer for part c.

03:14 Let's move on to part d.

03:16 So in part d, same exercise, but here they're giving us in part d a quotient f of x divided by g of x.

03:26 And we want to know what that derivative is when we evaluate at x equal to two.

03:33 So this is just simply going to use the quotient rule.

03:37 So f prime of x times g of x minus g prime of x, f prime of x, f of x, and f of x, and over g of x squared evaluated when x is 2 so let's just plug these values in i think we can do it directly so at 2 what is f prime of 2 f prime of 2 is 1 3rd so i've got 1 3rd g of 2 g of 2 is 2 minus g prime of 2 2 is negative 3 f of two f of two is eight and all of this over g of two squared so g of two is two square that and you get four okay so this is what one six so this is what one six plus 24 over four and so if we look at this so that is what, if we do the math on that, let's see.

05:06 What is one six? we can do, what is it? this is one over 24 plus six.

05:16 So that is, let me back up.

05:28 There's a little bit of a mistake here.

05:30 I ran into an error message, so real quickly, let me correct that.

05:34 We have the one -third times two.

05:38 One -third times two is two -thirds.

05:40 So sorry about that.

05:43 So this is two -thirds plus 24 over four.

05:47 So that becomes what two -twelths plus 24 -4 is six...