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Find an equation of the tangent line to the curve at the given point y=sqrt{x} (1,1)
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\[y=\sqrt(x) \]
Given point (1,1).
First step differentiate
We do this because we know f'(1) will give us the slope of the line that is tangent to the curve y=sqrt(x) at (1,1)
point slope form for a line that contains the point (a,b) is y-b=m(x-a) where m is the slope... so here you will have: y-f(1)=f'(1)(x-1) where f(1)=sqrt(1)=1 <--this was actually given so you have y-1=f'(1)(x-1) only work for you to do is to find f'(1) and replace f'(1) here
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Alright, thank you very much!
np
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