In Exercises 1-6, (a) express d w / d t as a function of t, both by using the Chain Rule and by

step 1 Here we have chapter 14, section 4, problem number 6. This is a part two part question where we'

In Exercises 1-6, (a) express d w / d t as a function of t,...

Key Concepts

Evaluating Derivatives at a Point

Evaluating Derivatives at a Point means substituting a specific value of the independent variable into the derivative after computation to determine the rate of change at that particular point. This process provides a numerical value that represents how the dependent variable changes with respect to the independent variable at that specific instance.

Function Composition

Function Composition refers to the process of combining two or more functions such that the output of one function becomes the input of the next. This concept is central to problems where variables are defined in terms of others, requiring the use of strategies like the chain rule, because the overall function’s derivative depends on the derivatives of its individual components.

Direct Differentiation

Direct Differentiation involves rewriting a composite function entirely in terms of the independent variable and then differentiating it in the usual way. This method avoids the layered complexity of the chain rule by expressing the function in a simplified single-variable form before applying standard differentiation techniques.

Chain Rule

The Chain Rule is a fundamental principle in calculus used to differentiate composite functions. It states that the derivative of a function that is composed of other functions is the product of the derivatives of the outer function evaluated at the inner function and the derivative of the inner function. This concept is critical when dealing with functions that depend on several intermediary variables, each of which is itself a function of an independent variable.

Transcript

00:01 Here we have chapter 14, section 4, problem number 6.

00:07 This is a part two part question where we're trying to find dwdt or the derivative of w with respect to t.

00:17 We can see here that w is z minus sine xy.

00:22 There are no t's.

00:23 So we probably want a formula.

00:26 So our formula for the derivative of w with respect to t it's going to be the partial derivative of w with respect to x times the derivative x with respect to t plus partial derivative w with respect to y times the derivative of y with respect to t plus partial derivative w with respect to z times derivative of z with respect to t right these can get pretty lengthy so here we're looking for partial derivative of w which we said was z minus sign x times y and that's going to be with respect to x times the derivative of x which is t plus partial derivative of z minus sine x times y with respect to y times the derivative of y which is natural log of t plus partial derivative of z minus sign x times y partial with respect to z times the derivative with respect to t of e to the t minus one.

01:44 Alright? so here we have z minus sine x y and we're only looking for the derivative with respect to x.

01:53 So z is a constant, that's going to be zero.

01:56 This negative sign x y, we're going to look at our center here first, x times y.

02:02 I'm going to go ahead and pull out that negative preemptively here.

02:05 So we have x y, derivative with to x if y is a constant and x is x x goes to one we can pull out that y and now we can look at the outside this sign through sign is cosine all right and the derivative with respect to t is just one all right so again now we're looking for y so that's going to be negative x cosine x y and during the derivative of the natural log of t is 1 over t.

02:50 And over here we have with respect to z, so all of this can just go away since x and y are constants.

02:58 So derivative of z is just 1.

03:01 All right? and our derivative of e to the t minus 1 is just e to the t minus 1, since the derivative of t minus 1 is just 1.

03:14 So what we're left with here is going to be negative.

03:18 Y cosine xy minus x cosine xy all over t plus e to the t minus one all right so at this stage everything is as simplified as we can get it for the time being and we have a y and an x here several of them and so we want to go ahead and get rid of all those so we're going to go ahead and input our values here so we had a negative y and y was the natural log of t so that's going to be negative natural log of t cosine x y becomes t natural log of t minus x was t cosine cosine t natural log of t all over t plus e to the t minus one right so we can see these ts are going to cancel t divided by t is just one and we can see that we can probably pull out this cosine, which negative cosine, negative cosine, t natural log of t.

04:31 And what's left over, it's going to be a natural log of t, and over here is just a 1.

04:36 All right.

04:38 And we have our e, t minus 1.

04:41 All right.

04:43 And that's it.

04:45 That's as simplified as we can get it.

04:49 So the second part of a says that we we should do this with respect to t directly.

04:58 So we have dwdt, which is the derivative of w with respect to t, and we have w equals z minus sign xy...