Posted by **Daniel** on Thursday, February 24, 2011 at 9:39pm.

-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?

- calculus -
**MathMate**, Thursday, February 24, 2011 at 9:56pm
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.

- calculus -
**Daniel**, Thursday, February 24, 2011 at 10:03pm
m=pi

and b= y-pi=pi(x+5pi)

y-pi=pi(x)+5pi(pi)

m=pie but what does b= though

- calculus -
**MathMate**, Thursday, February 24, 2011 at 10:47pm
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.

- calculus -
**Reiny**, Thursday, February 24, 2011 at 10:49pm
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

- calculus -
**Daniel**, Thursday, February 24, 2011 at 11:31pm
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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