f(x)=12x/(sinx+cosx)
find f'(-pie)
need some help!
use the quotient rule:
http://www.allaboutcircuits.com/vol_5/chpt_6/6.html
here are the derivatives of some elementary functions:
http://www.karlscalculus.org/divrules.html
Did you find f'(x) first?
I just read your earlier post and you said you plugged -pi into
12x/(sinx+cosx) .
You have to find f' first, then plug in (-pi).
The derivative, f' is,
(12(x sin x + sin x - x cos x + cos x)/
(sin x + cos x)^2
Then you plug in (-pi)
You realize that the derivative can be in various forms depending on how you factor it.
I also checked my answer in an online calculator and it said both answers are correct, the derivative and f'(-pi).
f'(-pi) = -12 - 12pi
or, 12(-1 - pi)
Where are you checking your answer?
Daniel, thought I would jump in here again.
The answer that Helper gave you several times is correct, I had the same result as did MathMate.
The problem appears to be the way you are expected to enter that answer in an answer window on some kind of webpage.
without doubt the answer is
-12 - 12π, or
-(12+π) or
-12π-12 or
-12(1+π) or
etc
If your computer does not accept any of these variations, then it is the fault of their program.
Thank you so much Reiny.
To find the derivative of the given function f(x), you can use the quotient rule. The quotient rule states that for a function u(x)/v(x), the derivative is given by:
(f(x))' = (u'(x)v(x) - v'(x)u(x))/[v(x)]^2
Applying the quotient rule to the given function f(x) = 12x/(sin(x) + cos(x)), we let u(x) = 12x and v(x) = sin(x) + cos(x).
Now, we need to find u'(x) and v'(x):
u'(x) is the derivative of u(x) = 12x, which is simply 12.
v'(x) is the derivative of v(x) = sin(x) + cos(x). Applying the chain rule, we have v'(x) = cos(x) - sin(x).
Now, we can substitute these values into the quotient rule formula:
(f(x))' = [(12)(sin(x) + cos(x)) - (cos(x) - sin(x))(12x)] / [sin(x) + cos(x)]^2
To find f'(-π), we substitute x = -π into the derivative formula:
(f(x))' = [(12)(sin(-π) + cos(-π)) - (cos(-π) - sin(-π))(12(-π))] / [sin(-π) + cos(-π)]^2
Now we need to simplify this expression:
sin(-π) = 0 and cos(-π) = -1, so we have:
(f(x))' = [(12)(0 + (-1)) - ((-1) - 0)(12(-π))] / [0 + (-1)]^2
= [(-12) - (-12π)] / 1
= -12 + 12π
Therefore, f'(-π) = -12 + 12π.
So, the derivative of f(x) evaluated at x = -π is -12 + 12π.