Question

find an equation of the tangent line to the graph of f at the given point.
f(x)=√3x^2-2) point(3,5)

Answer 1

you need the derivative as the first step

Answer 2

ok. brb. gonna get it

Answer 3

the derivative of \(\sqrt{g(x)}\) is
\[\frac{g'(x)}{2\sqrt{g(x)}}\]

Answer 4

in your case it is
\[\frac{6x}{2\sqrt{3x^2-2}}\]

Answer 5

replace \(x\) by 3 to find the slope
then use the all mighty point slope formula

Answer 6

i got (1/2)(3x^2-2)^-1/2 (6x)

Answer 7

luckily you get a nice rational number as an answer,

Answer 8

hold the phone
don't be a slave to the power rule. you are right of course, but writing the derivative that way is useless if you want to evaluate it at a number
you should recall how to take the derivative of the square root of something. while every one else taking the exam is rewriting in exponential notation etc etc it will be to your benefit to know
\[\huge\frac{d}{dx}[\sqrt{g(x)}]=\frac{g'(x)}{2\sqrt{g(x)}}\]

Answer 9

woah!

Answer 10

very simple
the derivative of the thing inside the radical goes on top, divide by twice the radical of the inside piece
instantly you get
\[f'(x)=\frac{6x}{2\sqrt{3x^2-2}}\]now you can find
\[f'(3)\] easilyy

Answer 11

with the exponential notation, you cannot evaluate the derivative

Answer 12

the final answer is supposed to look like 9x-5y-2=0

Answer 13

we now need \(f'(3)\) which is possible when written properly
\[f'(3)=\frac{6\times 3}{2\sqrt{3\times 3^2-2}}\]
\[=\frac{18}{2\sqrt{25}}=\frac{18}{10}=\frac{9}{5}\]

Answer 14

now you are 75% done
point slope formula gives
\[y-5=\frac{9}{5}(x-3)\]and it is algebra from here on in

Answer 15

don't forget two common and important derivatives, just memorize them
\[\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\] and by the chain rule
\[\frac{d}{dx}[\sqrt{g(x)}=\frac{g'(x)}{2\sqrt{g(x)}}\]