f'(x) = (7(x-6) - 7x)/(x-6)^2
=-42/(x-6)^2
then f'(9) = -42/9 = -14/3
Let f(x)=7x/x–6
Then f'(9)=
And after simplifying f'(x)=
3 answers
Reiny sorry about the question previous to this.. yea i got it before hand i should have said that on the boards.
I appears as though i am having trouble with the product and quotient rule and i definitely don't understand the chain.
here's another problem that i attempted and used quotient rule
Let f(x)=2/5x+7.
f'(x)=
(5x+7)-(2)(5)/ (5x+7)^2
i typed this in and got it wrong
well i typed (5x+7)-10/ (5x+7)^2
I appears as though i am having trouble with the product and quotient rule and i definitely don't understand the chain.
here's another problem that i attempted and used quotient rule
Let f(x)=2/5x+7.
f'(x)=
(5x+7)-(2)(5)/ (5x+7)^2
i typed this in and got it wrong
well i typed (5x+7)-10/ (5x+7)^2
for f(x)=2/5x+7
f'(x) = [(5x+7)(0) - 2(5)]/(5x+7)^2
= -10/(5x+7)^2
notice the derivative of 2, the top, is zero. You had it as (5x+7)*(1)
in a nutshell, if y = u/v
then y' = (vu' - uv')/v^2
for the "chain" rule, I will use an example
y = 7(4x^2 - 5x)^6
y' = 42(4x^2 - 5x)^5 * (8x-5)
in general if
y = a(anything)^n then
y' = an(anything)^(n-1)*(anything)'
in words:
the exponent times the coefficient in front*(the base),raised to the exponent reduced by 1, times the derivative of the base.
f'(x) = [(5x+7)(0) - 2(5)]/(5x+7)^2
= -10/(5x+7)^2
notice the derivative of 2, the top, is zero. You had it as (5x+7)*(1)
in a nutshell, if y = u/v
then y' = (vu' - uv')/v^2
for the "chain" rule, I will use an example
y = 7(4x^2 - 5x)^6
y' = 42(4x^2 - 5x)^5 * (8x-5)
in general if
y = a(anything)^n then
y' = an(anything)^(n-1)*(anything)'
in words:
the exponent times the coefficient in front*(the base),raised to the exponent reduced by 1, times the derivative of the base.