Find f^'(c) by forming the difference quotient...

Key Concepts

Difference Quotient

The difference quotient is an expression used to compute the derivative of a function at a specific point. It represents the average rate of change of the function over a small interval, given by [f(c+h) - f(c)]/h, and forms the foundational concept for defining derivatives as the limit of this expression as h approaches zero.

Derivative

The derivative of a function measures the instantaneous rate of change of the function with respect to its variable. It is defined as the limit of the difference quotient as the interval parameter h tends to zero, capturing the behavior of the function at an infinitesimally small scale.

Limit

The concept of limits in calculus is used to determine the value that a function approaches as the input approaches a particular point. When applying the difference quotient, the limit as h approaches zero is used to transition from an average rate of change to the precise instantaneous rate of change.

Power Rule

The power rule is a basic technique in differentiation, stating that the derivative of a function of the form x^n is n*x^(n-1). This rule simplifies the process of differentiating polynomial functions and is essential for quickly determining derivatives without having to use the formal limit definition every time.

Transcript

00:03 Here we have f of x equals 5 minus x to the fourth, and we want to find f prime of negative 1 by taking the limit as each approach is 0 of the difference quotient.

00:14 F of negative 1 plus h minus f of negative 1 divided by h.

00:21 So we're going to take the limit as h approach is 0 of f of negative 1 plus h.

00:34 F of x is 5 minus x to the 4th.

00:37 So f of negative 1 plus h, the negative 1 plus h is going to get substituted in for x.

00:44 So f of negative 1 plus h will be 5 minus x to the 4th, which will be negative 1 plus h to a 4th.

01:00 We have to be really careful that we don't make a little mistake.

01:05 Because if we make a little mistake now, it'll carry through the remainder of the problem, and we'll get the wrong answer.

01:12 Okay, so f of negative 1 plus h is 5 minus negative 1 plus h to the 4th.

01:20 Then we need to subtract f of negative 1.

01:25 Substituting negative 1 into the f of x function, we get 5 minus x to the 4th, which will be negative 1 to the 4th.

01:42 Now let's make this a square bracket just to make it a little bit easier to see.

01:47 So f of negative 1, 5 minus negative 1 to the 4th.

01:51 5 minus negative 1 to the 4th.

01:54 And then that all gets put over each.

02:05 All right.

02:06 This piece here, we can carefully and easily calculate this.

02:13 Negative 1 to the 4th is a positive 1.

02:16 So 5 minus a positive 1 is 4.

02:20 So this is going to be 4.

02:22 Keep in mind that'll be subtracted pretty soon but not yet what we need to do is we need to expand out this right here the negative 1 plus h to the 4th negative 1 plus h to the 4th if we expand it out we will get now just so you know i actually rewrote this negative 1 plus h is the same thing as h minus 1.

03:01 So when i actually expanded this out to, you know, to the fourth power, negative 1 plus h is h minus 1.

03:10 So i actually expanded h minus 1 to the fourth.

03:14 And when we do that, we get h to the 4th, minus 4 h cubed plus 6h squared, minus 4h plus 1.

04:01 Now, five is subtracting that quantity.

04:06 So five has to subtract that entire expression.

04:13 So let's bring down this page a little bit.

04:17 It gives a little bit of room.

04:21 Okay...