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how do i find an equation for the tangent to the graph y = arcsecx at x=2
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First find dy/dx = ???
Are you sure it's x=2?
yes thats what it says..
Oh... okay I was reading that wrong... so can you find dy/dx = ???
Do you need help finding the derivative?
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\[\frac{ 1 }{ \left| x \right| \sqrt{x^2-1}}\]?
yes
So you can find the slope using the derivative... Find the slope of the line at x=2
slope = \(\large m = \frac{ 1 }{ \left| 2 \right| \sqrt{(2)^2-1}}=\frac{1}{2\sqrt3} \)
Now all you need is the y-coordinate when x=2 So you'll then have a point and a slope to write the equation of the tangent line in point-slope form
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okay hang onn
well its the same thing lol
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