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(a) If $ F(x) = f(x)g(x). $ where $ f $ and $ g $ have derivatives of all orders, show that $ F" = f" g + 2f'g' + fg". $(b) Find similar formulas for $ F''' $ and $ F^{(4)}. $ (c) Guess a formula for $ F^{(n)}. $

(a) See explanation for proof(b) $F \prime \prime \prime =f \prime \prime \prime g+ 3 f \prime g \prime \prime+ f g^{\prime \prime \prime}$$F^{(4)}=f^{(4)} g+ 4f \prime \prime \prime g \prime+ 6f \prime \prime g \prime \prime+ 4f \prime g \prime \prime+ f g^{(4)}$(c) $F^{n}=\sum_{0}^{n} \frac{n !}{k !(n-k) !} f^{n-k} g^{k}$

00:54

Frank L.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 2

The Product and Quotient Rules

Derivatives

Differentiation

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

02:34

Assuming the first and sec…

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Find a general formula for…

00:42

Suppose $ f $ is different…

05:48

(a) Prove:$$\begin…

01:55

Let $f(x)$ have derivative…

03:11

Assume $f, g,$ and $h$ are…

04:54

Quotient Rule for the seco…

06:04

Assume that$$f(0)=2, \…

02:31

Find the $ n $ th derivati…

04:10

(a) Use the Product Rule t…

Hey, it's Claire. So ending. Read here. So for part A, we're going to use the first turret, the product roll, to find the first trip it is. And then we're going to use the product roll on both products to find the second derivative turns the derivative of G plus f of x times, a second or a bit of G of X, just equal to the second derivative of friends G of X plus two terms. The first derivative of terms derivative of G plus that F G comes the second derivative of G for part B, we have the equation for the second derivative. We're gonna difference you again to get the third derivative we got. Dream the second derivative drag this. And when we simplify it, we get three times the first derivative of the second derivative of G plus the terms such a threat of dirt derivative of G. We're gonna different. She again me yet I have to. The fourth is equal to this becomes equal Thio to the fourth three plus six back to the second derivative G to the second derivative plus four tongues, the derivative of G character of it. If and then for part C, we see that where we're going to recall the 2nd 3rd and fourth derivatives. And we see the coefficient. So for the half, or the coefficient of each derivative for the seconds we have 1 to 1. For the third, we have 1331 And for the fourth, we have one for six for one. So these are the paschal triangles. And you think if I know meal expression, this is our formula to the n minus K terms G to the key.

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