In Exercises 1 through 6, apply Definition 19.4 .1 to find each of the partial derivatives. f(x,

In Exercises 1 through 6, apply Definition 19.4 .1 to find...

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Key Concepts

Partial Derivative

A partial derivative measures how a multivariable function changes as one of its variables changes, while all other variables are held constant. It is fundamental in multivariable calculus and is used to analyze the local behavior of functions of several variables.

Limit Definition of a Partial Derivative

The limit definition of a partial derivative provides a rigorous mathematical formulation for finding the rate of change of a function with respect to one variable. This involves computing the limit of the difference quotient as the increment in that variable approaches zero, ensuring that the derivative captures the instantaneous rate of change.

Difference Quotient

The difference quotient is a key component in the definition of derivatives. For partial derivatives, it involves taking the difference between the function evaluated at a specific increment in one variable and the function evaluated at the original point, then dividing by the size of the increment. This quotient approximates the derivative, with the approximation becoming exact in the limit as the increment tends to zero.

Transcript

00:02 Okay, section 3 .2, function definition of a derivative problem number two, as many times as you can, you want to be grounded in this limit definition of the first derivative.

00:16 So the limit, h approaches 0, f of x plus h, minus f of x over h.

00:22 Okay.

00:23 We are presented with a function f of x equal x minus 1 squared plus 1.

00:34 Okay, and so we're asked to find its derivatives, so f of x minus one squared plus one.

00:40 So we've got a couple of choices here.

00:43 We could go ahead and distribute this out to make the algebra a little bit easier.

00:48 So that's just one option for doing this.

00:50 The other thing is just to plug in values for x plus h to make this happen.

00:55 So what i would tend to do is if i can simplify earlier in the process, i will do that.

01:00 So this is x squared minus 2x, and this is going to be plus 1 plus 1.

01:08 So this is f of x equal x squared minus 2x plus 2.

01:17 Okay, so that's just algebraically expanding that binomial, getting one expression.

01:22 So i know now that the derivative f prime of x will equal the limit as h approaches zero.

01:32 So f of x plus h, replace all those xs with x plus h's.

01:37 So x plus h squared, minus 2, x plus h plus 2, and then minus f of x.

01:48 So that's going to be minus x squared plus 2x minus 2, all of that over h...

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