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Aim: How do we differentiate using the product rule? Name: Section: Get Ready:
1. Find if
2. Find if
Evaluate y’ and y” at x=1 and x = 12. What does
the value of y’ and y” tell you about y at these values of x?
I. Product Rule
or if , then
1. Write the equation of the tangent line at x=1 and x=-‐1.
2. Write the equation of the tangent line at x=1.
f '(x)f (x) = 4x3 − 2
x3+ 6 x23 −
4x
y '' y = 3x3 + 4x2 − 5x+ 2
x2
ddx[ f (x) * g(x)] = f '(x)g(x)+ f (x)g '(x)
f = u *v f ' = u 'v+uv '
f (x) = (5x − 4x3 )(4x −1)
f (x) = (2x2 − 3x+ 2)2
Aim: How do we differentiate using the product rule? Name: Section:
3.
4. II. Find the derivative using two methods (one is the product rule)
1.
2.
3.
4.
f (x) = (4x2 − 2x+ 3)(−2x4 + x3 − 5x2 − x+1)
f (x) = (x2 + 5x − 2)(2x3 + 5x2 + 7x − 5)
y = (5x4 )(5x2 + 3x − 5)
f (x) = (5x4 )(5x2 + 3x − 5)
f (x) = (x6 + 4x)(3x3 − 4x+ 2)
s(t) = (2t − 5)(2t 2 − 9)+ (5t)(2t 3 − 3t)
Aim: How do we differentiate using the product rule? Name: Section:
5. III. Practice: Find the derivative of each function 1. 𝑦 = !
!𝑥 + 5
2. 𝑓 𝑥 = !
!− !
!!+ !
!!!
3. 𝑠 𝑡 = !!
𝑡! − 3𝑡! + 2𝑡 − 6 4. 𝑔 𝑥 = 4𝑥! − 1 2𝑥! − 3𝑥! + 6𝑥! + 9𝑥 − 7 5. 𝑑 𝑥 = 5𝑥! − 6𝑥! + 11 3𝑥! − 20
f (x) = (3x+ 2)(5x3 + 2x)+ (3x2 )(4x5 + x)
Aim: How do we differentiate using the product rule? Name: Section: 6. 𝑦 = !
! !− 4𝑥! + !
!+ !
! !!
7. 𝑦 = !
!− 2 𝑥! + 3𝑥 !
!!! + 5𝑥! − 2𝑥
8. Evaluate the derivative of 𝑓 𝑥 = !
!!3𝑥! − 6𝑥! + 2𝑥 − 11
using the product rule and evaluate the derivative at x = 1.
9. Write the equation of the tangent line to the
function at the given point.
𝑦 = 𝑥! + 3𝑥! − 21𝑥! − 3 𝑥 𝑎𝑡 𝑥 = 1