Find the value of the derivative (if it exists) at each indicated extremum. f(x)=-3 x √(x+1)

Name: Problem 4, Chapter 4: Calculus Early Transcendental Functions
Uploaded: 2021-07-17T00:26:49-08:00
Duration: 5 min 22 s
Channel: Jacquelyn Trost
Description: Find the value of the derivative (if it exists) at each indicated extremum. f(x)=-3 x √(x+1)

Find the value of the derivative (if it exists) at each...

Key Concepts

Derivative

The derivative is a fundamental concept in calculus that represents the instantaneous rate of change of a function with respect to its variable. It gives the slope of the tangent line to the function at any given point, which can be used to analyze how the function behaves—whether it is increasing, decreasing, or remaining constant.

Extrema

Extrema refer to the local maximum and minimum values that a function can attain. These are critical points where the function changes direction, and they are typically identified by setting the derivative equal to zero or by finding where the derivative does not exist.

Product Rule

The product rule is a method used in differentiation when a function is expressed as the product of two or more differentiable functions. It states that the derivative of the product is the first function times the derivative of the second plus the second function times the derivative of the first. This rule is essential when dealing with functions that are multiplicative combinations of simpler expressions.

Chain Rule

The chain rule is a key technique for differentiating composite functions. It allows one to differentiate a function that is nested within another function by multiplying the derivative of the outer function, evaluated at the inner function, by the derivative of the inner function itself. The chain rule is especially important when dealing with functions that include nested operations such as square root or exponentiation.

Differentiability

Differentiability refers to the existence of a derivative at a particular point of a function. A function may fail to be differentiable at points where it has a sharp corner, discontinuity, or vertical tangent. Assessing differentiability is especially important at extrema, as these points may be candidates for where the derivative is zero or does not exist, thereby affecting how the extrema are characterized.

Transcript

00:01 For this problem, we've been given the function f of x equals negative 3x times the square root of x plus 1.

00:07 And we need to evaluate the derivative of this function at two points.

00:13 The first point is negative 1 -0.

00:15 And the second point is negative 2 thirds, 2 square root of 3 over 3.

00:22 Now, in order to evaluate the derivative at these points, we first need to find what the derivative is.

00:28 Now before we do this, there's two pieces we need to review.

00:32 First, this function is a product.

00:34 I have negative 3x times that square root.

00:38 So let's review what the product rule looks like.

00:41 If i have two functions, let's say they're f of x and g of x and they're being multiplied together, and i want to find what that derivative is.

00:48 The product rule says i will take the first function times the derivative of the second plus the second function times the derivative of the first.

01:00 So that's the rule that we're going to need when we take this derivative.

01:04 The second thing i just want to review is taking the derivative of a square root.

01:09 So let's make it very simple.

01:12 Let's take, we'll just take like the square root of x to remind us how this works.

01:16 We can rewrite the square root of x as x to the one half.

01:22 So when we take the derivative, we bring down that exponent and then we subtract 1.

01:29 So we end up with 1 half times x to that negative 1⁄2.

01:33 And we could rewrite that to look like that.

01:39 We could put that square root into the denominator.

01:41 We've kind of written it as a fractional exponent and then changed it back to a square root.

01:46 Because while the fractional exponent makes it easier to take the derivative, this is often an easier form to see exactly what's going on as the fraction.

01:54 You're much less likely to end up accidentally dividing by zero because you can see that it's in the denominator there.

02:00 So we'll use both of those pieces when it comes to taking this derivative.

02:06 So first, product rule, the first function, which is negative 3x, times the derivative of the second.

02:16 Well, that's that square root.

02:17 So if you remember what we just looked at here, that will end up putting a two from that fractional exponent, and the square root in the denominator.

02:27 Now, i would also then, which we didn't do on that green piece, was if it wasn't just x, if it's a function of x, you would use the chain rule.

02:34 So we'd have to multiply by the derivative of what's under that exponent, but the derivative of x plus 1 is just 1.

02:40 So we are good just like this...

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