find an equation of the tangent line to the curve at the given point. y=sqrt(x) (1,1) Ive never been good with these equations and the square root is confusing me.

Question

find an equation of  the tangent line to the curve at the given point.  y=sqrt(x)   (1,1)  Ive never been good with these equations and the square root is confusing me.

anonymous · Answer

you should memorize the derivative of $f(x)=\sqrt{x}$ because 
1) it is a very common function and comes up frequently
2) math teachers love to use it on tests and 
3) it never changes so if you memorize it you will save a lot of time on quizzes, homework, and most importantly exams

anonymous · Answer

just like you know $8\times 7=56$ you should know 
$$\frac{d}{dx}[\sqrt{x}]=\frac{1}{2\sqrt{x}}$$

anonymous · Answer

find y' = slope of tangent line to the curve = 1/ 2* sqrt(x)
slope at point (1,1) = 1/2*1=0.5
eqn of line is y-y1=m*(x-x1
y-1=0.5(x-1)
or 2y-2=x-1,

anonymous · Answer

now that you have memorized it, the slope is easy to find. 
$$f(x)=\sqrt{x},f'(x)=\frac{1}{2\sqrt{x}},f'(1)=\frac{1}{2\sqrt{1}}=\frac{1}{2}$$

anonymous · Answer

you got the slope, it is $\frac{1}{2}$ and you have the point, it is $(1,1)$ point slope formula does the rest

laddiusmaximus · Answer

how does that simplify?

anonymous · Answer

@LaddiusMaximus , I already gave the answer to you

laddiusmaximus · Answer

i know. I just dont understand how you came to it.

anonymous · Answer

whats confusing you, just explain

laddiusmaximus · Answer

how does y-y1=1/2(x-x1) become 2y-2=x-1?

anonymous · Answer

(x1, y1) is the point  where 1 is subscript

the coordinates of the point are given to us (see the question) which are (1,1)
so  y-1=0.5(x-1)

anonymous · Answer

does that help

laddiusmaximus · Answer

yes but then you get 2y-2=x-1  thats where Im lost

anonymous · Answer

0.5=1/2
so 2(y-1)=1(x-1)

anonymous · Answer

does that make sense ?

laddiusmaximus · Answer

yes. I kept moving the entire 1/2 around