Let f(x)=7x/x–6
Then f'(9)=
And after simplifying f'(x)=
f'(x) = (7(x-6) - 7x)/(x-6)^2
=-42/(x-6)^2
then f'(9) = -42/9 = -14/3
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
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.
To find the derivative of the function f(x) = 7x / (x – 6), we can use the quotient rule, which states that if we have a function h(x) = g(x) / j(x), where g(x) and j(x) are functions of x, the derivative of h(x) is given by:
h'(x) = [g'(x) * j(x) - g(x) * j'(x)] / [j(x)]^2.
Let's apply the quotient rule to find f'(x):
First, let's find g(x) and j(x):
g(x) = 7x
j(x) = x – 6.
Now, we need to find the derivatives of g(x) and j(x):
g'(x) = d/dx (7x) = 7, since the derivative of x with respect to x is 1.
j'(x) = d/dx (x – 6) = 1, since the derivative of a constant (-6) with respect to x is 0, and the derivative of x with respect to x is 1.
Now we can substitute these values into the quotient rule formula:
f'(x) = [g'(x) * j(x) - g(x) * j'(x)] / [j(x)]^2
= [7 * (x – 6) - 7x * 1] / (x – 6)^2
= (7x – 42 - 7x) / (x – 6)^2
= -42 / (x – 6)^2.
Now, to find f'(9), we substitute x = 9 into the expression we derived for f'(x):
f'(9) = -42 / (9 – 6)^2
= -42 / 3^2
= -42 / 9
= -14.
After simplifying f'(x), we found that f'(x) = -42 / (x – 6)^2.