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Mathematics 6 Online
OpenStudy (anonymous):

Need explanation pleaase! Find the equation of the tangent line to the given curve at the given point. Can someone please explain?

OpenStudy (anonymous):

\[y = 1/cosx - 2cosx \]

OpenStudy (roadjester):

To find the tangent line, you must first find the derivative.

OpenStudy (anonymous):

given point is ( pi/3 , 1) and answer is 3\[3\sqrt{3x}-y+1-\pi \sqrt{3}=0 \]

OpenStudy (roadjester):

Then use the point-slope formula

OpenStudy (anonymous):

Can you show me?

OpenStudy (roadjester):

I'm sorry, I have to go...I'll be free tomorrow.

OpenStudy (experimentx):

find the derivative, put the values, you will get the slope since you have point and slope ... use single point slope formula

OpenStudy (anonymous):

Okay thanks:D but the problem is I keep getting the answer wrong.

OpenStudy (anonymous):

y = 1/cosx−2cosx y = - sinx/ cox²x + 2 sinx

OpenStudy (anonymous):

f (x) = - tanx/ cosx + 2sinx f ( pi/3) = -2√3 + √3 = - √3

OpenStudy (anonymous):

Tangent line at ( pi/3 , 1) : y - 1 = -√3 ( x - pi/3) => y = -√3 x + pi/√3

OpenStudy (anonymous):

@bobobo Not sure if your post miss some parentheses!

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