Question

Find the equation of the tangent line to the curve y =√x at x = 9. (that's a square root symbol over the x)

Answer 1

Find the derivative and evaluate at x = 9

Answer 2

the derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\) replace \(x\) by 9

Answer 3

if it helps, think of sqrt(x) as x^(1/2) and apply the power rule to find the derivative.

Answer 4

That's gives the slope of the tangent line, the tangent line should touch the curve at x=9, so we get:
\[ y(9)=√9=a9+b \Rightarrow b=3-9a \]
Where \(a=f'(9)\) is the slope and \(b\) is a constant, the line is therefore:
\[ y=ax+b=f'(9)x+2-9f'(9) \]
Now just find f'(9).

Answer 5

i don't

Answer 6

the derivative of \(\sqrt{x}\) is always \(\frac{1}{2\sqrt{x}}\) it never changes
just like \(8\times 7=56\) just know it

Answer 7

also \(\frac{1}{2}x^{-\frac{1}{2}}\) is useless if you want to actually compute anything

Answer 8

@satellite73 you would obviously simplify it further to \[\frac{ 1 }{ 2\sqrt(x) }\]

Answer 9

it is good to know how to get to the solution without "just knowing it".. in time, it will become a habit though, like you said. :)

Answer 10

you get the solution by computing once
never again
might as well use
\[\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\] get \(\frac{1}{2\sqrt{x}}\) and never look back

Answer 11

it's silly to think that one will memorize all the derivatives.

Answer 12

@satellite73 “I don't need to know everything, I just need to know where to find it, when I need it”
― Albert Einstein