Posted by Daniel on Thursday, February 24, 2011 at 9:39pm.
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:
(y-y0)=m(x-x0)
Substitute values and simplify to get the equation of the line.
m=pi
and b= y-pi=pi(x+5pi)
y-pi=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:
(y-y0)=m(x-x0)
(y-(-5π))=(-5)(x-π)
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(x-π)
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
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