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equation of tangent line to y=xcosx at x = pi
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substitute pi into equation and got y1 = 0
f ' (x) = cosx + x - sinx
which slope equals -pi
y-0 = -pi(x-pi) ?
f'(x) = cos(x) + -xsin(x) I believe
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no its not.
i thought so too but according to mathway.com its cosx - sinx
Your slope is -1.
can you show me how ?
cos(pi) =-1 sin (pi)=0 thus -1 +0 = -1
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its -xsinx
-x sinx same thing would be pi times sin(x) or pi times 0.
oh right.
okay so for the equation
y1 = 0 right ?
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y-xcosx= 1(x-pi) Now solve for y.
its negative 1 not positive 1
Right
so is it-1 - pi ?
y-xcos(x)=-1(x-pi) Solve for y.
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y = -x ?
OK solve the right side of the equation first. -1(x-pi) = what>
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