Find the tangent line and the normal line? o.o

Question

babynini · Answer

attachment

babynini · Answer

@iambatman

anonymous · Answer

The slope of the tangent line to the given curve at the given point is given by the value of the derivative at that point, so first thing to find is the derivative:
$$y=\frac{\sqrt x}{x+9}~~\implies~~y'=\frac{\dfrac{x+9}{2\sqrt x}-\sqrt x}{(x+9)^2}$$
What's the value of $y'$ at the point $(1,0.1)$? (Plug in $x=1$.)

jim_thompson5910 · Answer

You're probably familiar with tangents at this point (based on what I saw in your other questions). The normal is probably what you're not familiar with? If not, then the normal line is simply the line perpendicular to the tangent line and it goes through the same tangential point

babynini · Answer

Right, i'm not really sure what they're asking for o.o

jim_thompson5910 · Answer

do you see how @SithsAndGiggles got dy/dx ?

anonymous · Answer

you got this or no?

babynini · Answer

[sorry I was working on another one, but I am back!]
Em i'm a little confused. Just a moment, let me work through it.

anonymous · Answer

ok if you get stuck let me know
the idea is to find the derivative, then replace $x$ by $1$ to find the slope

babynini · Answer

[sorry I was working on another one, but I am back!]
Em i'm a little confused. Just a moment, let me work through it.

babynini · Answer

So the value of y' = 1/25 at (1,0.1) correct?

babynini · Answer

@zepdrix :)

babynini · Answer

@iambatman