The Product Rule Jordan Hammond. Definition/Steps to Solving The Product Rule takes the derivative...
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![Page 1: The Product Rule Jordan Hammond. Definition/Steps to Solving The Product Rule takes the derivative of a product. It is used with equations with two or.](https://reader036.fdocuments.us/reader036/viewer/2022082613/5697c01a1a28abf838ccec38/html5/thumbnails/1.jpg)
The Product RuleJordan Hammond
![Page 2: The Product Rule Jordan Hammond. Definition/Steps to Solving The Product Rule takes the derivative of a product. It is used with equations with two or.](https://reader036.fdocuments.us/reader036/viewer/2022082613/5697c01a1a28abf838ccec38/html5/thumbnails/2.jpg)
Definition/Steps to SolvingThe Product Rule takes the derivative of a product. It
is used with equations with two or more functions, such as y = f(x)g(x)
So, in order to find the derivative of y = f(x)g(x), you use this formula:
y1 = f(x)g1(x) + f1(x)g(x)
In simpler terms,
y1 = (First term)(Derivative of the Second term) + (Derivative of the First term)(Second term)
F(ds) + dF(s)
![Page 3: The Product Rule Jordan Hammond. Definition/Steps to Solving The Product Rule takes the derivative of a product. It is used with equations with two or.](https://reader036.fdocuments.us/reader036/viewer/2022082613/5697c01a1a28abf838ccec38/html5/thumbnails/3.jpg)
Example #1y = (x2)(x + 3x2)
y1 = F(ds) + dF(s)
= (x2)(6x+1) + (2x)(x+3x2)
= 6x3 + x2 + 2x2 + 6x3
= 12x3 + 3x2
y1 = 12x3 + 3x2
![Page 4: The Product Rule Jordan Hammond. Definition/Steps to Solving The Product Rule takes the derivative of a product. It is used with equations with two or.](https://reader036.fdocuments.us/reader036/viewer/2022082613/5697c01a1a28abf838ccec38/html5/thumbnails/4.jpg)
Example #2y = (ex)(x3+4)
y1 = F(ds) + dF(s)
(The Derivative of ex is ex)
= (ex)(3x2) + (ex)(x3 + 4)
y1 = (ex)(3x2) + (ex)(x3) + 4ex
![Page 5: The Product Rule Jordan Hammond. Definition/Steps to Solving The Product Rule takes the derivative of a product. It is used with equations with two or.](https://reader036.fdocuments.us/reader036/viewer/2022082613/5697c01a1a28abf838ccec38/html5/thumbnails/5.jpg)
Your Turn! Problem #1Find the derivative of the function:
y = (2x)(4x4 + 3x)
![Page 6: The Product Rule Jordan Hammond. Definition/Steps to Solving The Product Rule takes the derivative of a product. It is used with equations with two or.](https://reader036.fdocuments.us/reader036/viewer/2022082613/5697c01a1a28abf838ccec38/html5/thumbnails/6.jpg)
Go Again! Problem #2Find the derivative of the function:
y = (3ex)(9x)
![Page 7: The Product Rule Jordan Hammond. Definition/Steps to Solving The Product Rule takes the derivative of a product. It is used with equations with two or.](https://reader036.fdocuments.us/reader036/viewer/2022082613/5697c01a1a28abf838ccec38/html5/thumbnails/7.jpg)
Solution to Problem #1y = (2x)(4x4 + 3x)
y1 = F(ds) + dF(s)
= (2x)(16x3+3) + (2)(4x4+3x)
= 32x4+6x+8x4+6x
= 40x4+12x
y1 = 40x4+12x
![Page 8: The Product Rule Jordan Hammond. Definition/Steps to Solving The Product Rule takes the derivative of a product. It is used with equations with two or.](https://reader036.fdocuments.us/reader036/viewer/2022082613/5697c01a1a28abf838ccec38/html5/thumbnails/8.jpg)
Solution to Problem #2 y = (3ex)(9x)
y1 = F(ds) + dF(s)
= (3ex)(9) + (3ex)(9x)
= 27ex + 27xex
y1 = 27ex + 27xex