Taking the derivative of products, 𝑦=(7 𝑥+5 )2

¿Feb. 17, simplify first.

𝑦=49𝑥2+70 𝑥+25𝑑𝑦𝑑𝑥

=98 𝑥+70

Feb. 20,Power rule, chain rule.

𝑦=(7 𝑥+5 )2

𝑑𝑦𝑑𝑥

=2 (7𝑥+5 )2− 1 (7+0 )

𝑑𝑦𝑑𝑥

=2[ (7 𝑥+5 ) (7 )]𝑜𝑟 98𝑥+70

Quadratic, tangent slopes will not be the same for all x ε R

Product of linear functions

)

𝑑𝑦𝑑𝑥≠ (7 ) (7 )

Does the derivative depend upon the derivative of the individual linear functions?

YES

But don’t just take the derivative of each factor.

Using the “Product rule”

Derivative of the 1st factor

2nd factor Derivative of the 2nd factor

1st factor

h (𝑥 )=[ 𝑓 (𝑥 ) ] [𝑔 (𝑥 )]

The derivative of h(x) with respect to “x” is

=

the derivative of the first factor,

[ 𝑑 𝑓 (𝑥 )𝑑𝑥 ]

times the second factor

[𝑔 (𝑥) ] + [ 𝑑𝑔 (𝑥)𝑑𝑥 ]

plus the derivative of the second factor, times the first factor.

[ 𝑓 (𝑥 )]

The first factor The second factor

Differentiation using the product rule

Examples:

h (𝑥 )=[ 𝑓 (𝑥 ) ] [𝑔 (𝑥 )]Given: then

1. 𝑦=(4 𝑥3−5𝑥2+𝑥−2)(3𝑥+1)

𝑑𝑦𝑑𝑥

=¿ [ ] [ ] + [ ] 3 𝑥+1 3[ ] 4 𝑥3−5 𝑥2+𝑥−2

If you were to simplify the derivative, you would drop brackets using multiplication and then collect like terms. The textbook simplifies their answers.

𝑑𝑦𝑑𝑥

=¿

2. 𝑦=(3 𝑥+5 ) √ (2𝑥+5 )3𝑜𝑟 𝑦= (3 𝑥+5 ) (2 𝑥+5 )32

[ ] [ ] + [ ] [ ]3 32

(2 𝑥+5 )12 (2 ) 3 𝑥+5

Now simplify the derivative

3 or or𝑑𝑦𝑑𝑥

=15 (√2 𝑥+5 ) (𝑥+2 )

32−22=12