Need explanation pleaase! Find the equation of the tangent line to the given curve at the given point. Can someone please explain?
\[y = 1/cosx - 2cosx \]
To find the tangent line, you must first find the derivative.
given point is ( pi/3 , 1) and answer is 3\[3\sqrt{3x}-y+1-\pi \sqrt{3}=0 \]
Then use the point-slope formula
Can you show me?
I'm sorry, I have to go...I'll be free tomorrow.
find the derivative, put the values, you will get the slope since you have point and slope ... use single point slope formula
Okay thanks:D but the problem is I keep getting the answer wrong.
y = 1/cosx−2cosx y = - sinx/ cox²x + 2 sinx
f (x) = - tanx/ cosx + 2sinx f ( pi/3) = -2√3 + √3 = - √3
Tangent line at ( pi/3 , 1) : y - 1 = -√3 ( x - pi/3) => y = -√3 x + pi/√3
@bobobo Not sure if your post miss some parentheses!
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