(a) 5 f(x)-g(x), x=1 (b) f(x) g^3(x), x=0 (c) (f(x))/(g(x)+1), x=1 (d) f(g(x)), x=0 (e)

(a) 5 f(x)-g(x), x=1 (b) f(x) g^3(x), x=0 (c)...

Key Concepts

Exponentiation of Functions

Exponentiation of functions involves raising the output of a function to a power. In expressions like g(x)^3, the value produced by g(x) is raised to the third power. Similarly, negative exponents, such as (g(x) + f(x))^(-2), indicate the reciprocal of the squared value of the function combination. This operation is important for modeling and solving problems where polynomial behavior or inverse relationships are involved.

Function Composition

Function composition is the process of applying one function to the result of another. In mathematical notation, the composition f(g(x)) means first evaluating g(x) and then using its result as the input for f(x). This concept is crucial because it allows for the creation of complex functions from simpler ones and is widely used in various fields of mathematics and applied sciences.

Function Evaluation

Function evaluation is the process of determining the output of a function for a given input value. This is typically done by substituting the input value into the function's expression and simplifying the result. Evaluating functions at specific points is a fundamental task in most areas of mathematics and is essential for understanding the behavior of functions at particular arguments.

Arithmetic Operations on Functions

Arithmetic operations on functions involve combining two or more functions using operations such as addition, subtraction, multiplication, and division. These operations allow us to create new functions from existing ones by performing basic arithmetic on their outputs. For example, scaling a function by a constant or adding two functions together helps explore relationships and modifications in their behavior.

Transcript

00:01 In this question, where we're talking about the chain rule, we will be finding the derivatives of each of these expressions, given the numerical values listed here in the top right corner.

00:19 First of all, we're going to find the derivative of 5x minus g, or sorry, 5f minus g, where f and g are functions of x, at the x value 1, positive 1.

00:37 Note that all of these derivatives occur, we're going to evaluate at either 1 or 0, because we're only given numerical values at 1 and 0.

00:53 So first of all, we differentiate.

00:57 So i'm going to call the derivative capital d, just to give it some sort of name.

01:03 It doesn't mean anything, it's not a common notation, it's just what i want to call it here.

01:09 This derivative by the constant multiple and some difference rule is going to be 5 times the derivative of f minus the derivative of g.

01:26 And that is because it is the derivative of 5 times f minus g.

01:33 Now if we evaluate this at x equals 1, then we have 5 times f prime at x equals 1 minus g prime at x equals 1.

01:55 We know f prime at 1 is this, and g prime at 1 is this.

02:01 So we substitute the minus 1 third and the minus 8 thirds.

02:21 This leads us to minus 5 thirds plus 8 thirds, which is 1.

02:33 Next we have part b, where the function is f times g cubed.

02:39 And we're differentiating this and evaluating at x equals 0.

02:45 So again, i'm going to use d to represent that.

02:48 Derivative.

02:51 And this time since we have a multiple of functions, we're going to have to use more complicated differentiation rules, such as the product rule and the power rule, or the power chain rule for functions.

03:06 The product rule says the derivative of this product of two functions is the derivative of the first function times the second, plus the first function times the derivative.

03:24 Of the second.

03:26 Now the derivative of the second, which is g cubed, is going to require the use of another differentiation rule.

03:36 We know the usual rule for powers is going, or tells us to multiply the exponent onto the coefficient and decrease the exponent by one.

03:50 That's the power rule for powers of x, but since this is a function of x, we also multiply that by the derivative of the the base of the power, the thing that's being exponentiated.

04:12 Or we could also think of it as a composite function.

04:16 It's the function g cubed, or sorry, the function g inside the function and the cubed function.

04:24 We take the derivative of the outside and multiply that by the derivative of the inside.

04:37 Now that we have our derivative of the expression, we evaluate this at x equals 0.

04:50 Again, we get our values from the table up here.

04:56 F prime is equal to 5 at x equals 0.

05:00 G is equal to 1.

05:04 F is equal to 1 as well.

05:13 G is equal to 1, like we already saw.

05:18 And g prime is equal to 1 3rd.

05:25 So this is just the 3 that was there before.

05:29 We multiply that by 1, and then by 1 3rd.

05:35 And so we get 5.

05:38 Plus 1, which is 6.

05:43 Now what about part c? the expression is f over g plus 1, and we are evaluating the derivative at x equals 1.

05:58 So in this case, when we differentiate, we're going to encounter the quotient rule, and that says for the numerator of your derivative, you have the derivative of the top of the function you're differentiating, i mean, time the bottom minus the top times the derivative of the bottom, and then the bottom of your derivative is just the bottom of the original expression squared.

06:45 Okay, we have that.

06:47 Now we're evaluating at x equals 1.

07:04 So we need f prime and it g.

07:08 F prime at 1 is negative 1 3, g is minus 4.

07:26 Then we need f and g prime...

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