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

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

Key Concepts

Constant Multiple Rule

This rule states that if a function is multiplied by a constant, the derivative of the product is the constant multiplied by the derivative of the function. It is one of the basic linearity properties of differentiation and is useful when handling expressions where functions are scaled by constant factors.

Sum Rule

The sum rule explains that the derivative of the sum (or difference) of two functions is the sum (or difference) of their derivatives. This property is essential when differentiating expressions that are composed of several additive terms.

Product Rule

The product rule is used to differentiate the product of two functions. It states that the derivative of a product is given by the first function times the derivative of the second, plus the second function times the derivative of the first. This rule is crucial when functions are multiplied together and their interactions need to be considered.

Quotient Rule

The quotient rule provides a method to differentiate a function that is the ratio of two differentiable functions. It involves taking the derivative of the numerator and denominator and combining them in a specific formula to account for their division, thus capturing the change in the ratio.

Chain Rule

The chain rule is applied when differentiating a composite function—that is, a function that is nested inside another function. It establishes that the derivative of the composite function is the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function, allowing for the systematic breakdown of complex derivative problems.

Power Rule

The power rule is a basic differentiation rule used to find the derivative of a function raised to a power. It states that the derivative of x raised to a constant exponent is the exponent multiplied by x raised to one less than that exponent, and it is often used in conjunction with the chain rule for more complex expressions.

Transcript

00:01 All right, we've been given this table of values.

00:04 We know what, when x is 2, we know the value of f of x, g of x, and both of the derivatives.

00:11 And when x is 3, we know the value of f of x, g of x, and both of the derivatives.

00:15 And we're being asked to find the value of each derivative with respect to x at a given x value.

00:21 So in part a, we have two times f of x that we want to find the derivative of when x is equal to 2.

00:29 So this is pretty straightforward here.

00:32 Because we have a constant, it's just going to be two times the derivative of x.

00:37 So it's going to be two times the derivative of f prime at 2.

00:44 And f prime at 2 is equal to 1 3rd.

00:47 So this just comes out to be 2 times 1 3rd, which is 2 thirds.

00:54 In part b, we have f of x plus g of x as our function.

01:01 And we want to find the value when x is equal to three of its derivative.

01:06 So when we are adding, we can take the derivative of each of them separately.

01:15 So this is just going to be f prime of x plus g prime of x.

01:20 When we plug in a three here, f prime of three plus g prime of three, is going to equal 2 pi plus 5.

01:39 Moving on to part c, we want the derivative of f of x times g of x.

01:47 So when we have the product of these functions, we are going to do the derivative of the first times the second, plus the first times the derivative of the second.

02:05 We're plugging in a 3 here.

02:08 So we have f prime of 3 times g of 3 plus f of 3 times g prime of 3.

02:17 So from our table, f prime of 3 is 2 pi.

02:23 G of 3 is negative 4.

02:28 We're going to add that to f of 3, which is equal to 3, times g prime of 3, which is 5.

02:37 So this comes out to be 45 minus 8 pi.

02:49 I'm going to copy this table so that we can see it here later as we move it down.

02:55 Let's copy it and paste it as an image.

02:59 We're just going to move it around down here.

03:01 So in part d, we are asked to find the derivative of f of x divided by g of x when x is 2.

03:15 So this is going to be that quotient rule.

03:18 So we do low times the derivative, high, low, d high, minus high, d low, so f of x times, times g prime of x over low squared so over g of x squared we're plugging in a two here so g of two is it's a highlight this so we don't get confused g of two is two times f prime of two is one third minus f of two is eight times g prime of two is negative three so there's our numerator over g prime i'm sorry not g prime over g squared so two squared so we have two -thirds plus 24 over four so 24 and two -thirds is 74 over 16 i wrote a 4 instead of a 3.

04:35 That's over 3.

04:38 So that's going to be 74 over 12.

04:42 Ooh, glad i caught that.

04:46 Double check yourself, especially with these tables.

04:48 My eyes go all over the place.

04:53 Let's take a look at part e.

04:55 In part e, this is the chain rule.

04:58 We want to find f of g of x when x is equal to 2.

05:04 So this is going to be the derivative of, so this is going to be f prime of g of x times the derivative of the inside, so times g prime of x.

05:22 So this one, to evaluate, we need to find f prime of g of 2 times g prime of 2.

05:31 So, g of 2, according to our table, is 2.

05:37 So f prime of 2 times g prime of 2.