calculus
posted by Daniel .
find the equation of the tangent line to the curve y=5xcosx at the point (pi,5pi)
the equation of this tangent line can be written in the form y=mx+b where
m=
and b=
what is the answer to m and b?

f(x)=5xcos(x)
f'(x)=5cos(x)5xsin(x)
=5(cos(x)xsin(x))
m=f'(π)
and the equation of the tangent passing through the point (x0,y0)=(π, 5π) is:
(yy0)=m(xx0)
Substitute values and simplify to get the equation of the line. 
m=pi
and b= ypi=pi(x+5pi)
ypi=pi(x)+5pi(pi)
m=pie but what does b= though 
Given
(x0,y0)=(π, 5π)
We have established:
m
=f'(π)
=5(cos(π)π*sin(π))
=5(1  π*0)
=5
The equation of a line of slope m passing through (x0,y0) is:
(yy0)=m(xx0)
(y(5π))=(5)(xπ)
Simplify to get the required equation.
b=0 by coincidence.
See graph:
http://img10.imageshack.us/i/1298601541.png/
Check my work. 
you did not follow MathMate's suggestion
f'(x) = 5(cosx  xsinx)
f'(π) = 5(cosπ  πsinπ)
= 5(1  π(0)) = 5 , so m = 5
equation:
y = 5x + b , but (π,5π) lies on it, so
5π = 5(π) + b
b = 0
equation : y = 5x
or using MathMate's suggestion:
y + 5π = 5(xπ)
y + 4π = 5x + 5π
y = 5x 
ok how about this one:
f(x)=12x/sinx+cosx
find f'(pie)
this is what i did:
12(pi)/(0+(1))
= 12pi
is that right what am i doing wrong and what would be the answer