1) Find the equation of the tangent line to the curve y= 6sec(x)-12cos(x) at the

Question

1) Find the equation of the tangent line to the curve y= 6sec(x)-12cos(x) at the point (pi/3,6). Write the answer in y=mx+b form (where m is slope and b is y intercept).

2) Find the points on the curve y=cos^2(x) that have horizontal tangent.

Please show all work.

Also, if you could clarify something for me I would appreciate it. When I work the problem f(x)=tan(x)cot(x) I got cot(x)*(sec(x))^2-(csc(x))^2*tan(x) but when I type it into wolfram alpha it says the derivative is zero. Am I right or are they? I see how both could be correct.

Thanks!!

Explanation / Answer

1.

y'=6*sec x*tan x+12*sin x

at x=pi/3

y'=31.177

so equation:

y-6=31.177*(x-(pi/3))

Q.

y'=2*cos(x)*(-sin(x)=-sin(2*x)

horizontal tangent means y'=0

so 2*x=n*pi

where n=0,1,2,3,...etc

note:

tanx*cot x=1

f(x)=1==> f'(x)=0

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