Question

HELP? pleaseeee.
use the quotient rule to differentiate the function…..

f(t)=cost/t^3

Answer 1

Easy

it's (cost) / t^3 right?

(cost)(t^-3)
= ((-sint)(t^-3) + (cost)(-3t^-4)) / (t^-3)^2
= ((-sint)(t^-3) + (cost)(-3t^-4)) / (t^-6)

Answer 2

yessss, why -3 though?

Answer 3

yes its [ (-sin t) * t^3 - cos t * 3t^2 ] / t^6

Answer 4

ohhh ok ok

Answer 5

because the power rule brings t^-3 to -3t^-4.

t^-3 = 1/t^3

Answer 6

I also nee help with this one….
- find f'(x) and f'(c )

f(x)=(x^2-2x+1)(x^3-1) c=1

Answer 7

u can simplify

t^2[- t sint - 3 cos t] / t^6

= [-t sint - 3cost] / t^4

Answer 8

f(x)=(x^2-2x+1)(x^3-1)
use the product rule
f'(x) = 3x^2(x^2 - 2x + 1) + (x^3 - 1)(2x - 2)

f'(c) = 3(1 - 2 + 1) + (1 - 1)(2 - 2) when c = 1
= 3(0) + 0 = 0

Answer 9

thankyou so much. I actually get how you got that now:) thanks!

Answer 10

@cwrw238