Question

Let f(x)=sqrt(4x-3) and g(x)=f(x)/x
What is the slope of the graph of f at x=3?

Answer 1

Does that mean I have to find the derivative?

Answer 2

yes

Answer 3

\[f(x)= \sqrt{4x-3}\]
\[g(x)=f(x)/x\]

Answer 4

so is it zero?

Answer 5

what is f'(x) ?

Answer 6

2x^-1/2

Answer 7

not quite

Answer 8

but do I have to plug in the 3

Answer 9

first?

Answer 10

sqrt(u) to u' / 2sqrt(u)

Answer 11

u' = 4 soo

f' = 2/sqrt(4x-3)

Answer 12

wait, what? what is 'sqrt(u) to u' / 2sqrt(u)'

Answer 13

you know what a chain rule is right?

Answer 14

nope :( I don't think I'm there yet

Answer 15

well then youll have to work it with th elimit rule then

Answer 16

I only know about the slope of tangent line (kinda-ish) the quotient and the power rule

Answer 17

if we have a composition of fuctions, one function inside of another, then we apply what is called a chain rule

h(t) derives to h'(t(x)) * t'(x) * x'

but by convention: x' = dx/dx = 1

Answer 18

right now we have a sqrt function with a linear function side of it

sqrt(u) , such that u = 4x-3

the derivative is therefore: [sqrt(u)]' * u'

Answer 19

*inside of it

Answer 20

so \[\sqrt{4x-3} * 4\] ?

Answer 21

almost, the first part needs a derivative

Answer 22

\[\frac d{dx} \sqrt{4x-3}\]
\[\frac d{dx} (4x-3)^{1/2}\]
well, if this was sqrt(x) we would be able to define it, so lets define the innards by 1 variable, let u = 4x-3
\[\frac d{dx} (u)^{1/2}=\frac{du}{dx}\frac{1}{2u^{1/2}}\]
but what is du/dx? we solve that from our definition of u
\[\frac d{dx}u = \frac d{dx}4x-3\]\[\frac {du}{dx} = \frac {dx}{dx}4\]
therefore du/dx=4
\[\frac d{dx} (u)^{1/2}=\frac{4}{2u^{1/2}}\]

Answer 23

So I get \[8x^{-1/2}\]

Answer 24

\[\frac{4}{2u^{1/2}}=2u^{-1/2}=2(4x-3)^{-1/2}\]

Answer 25

let x=3
\[2(4(3)-3)^{-1/2}\]
\[2(9)^{-1/2}\]
\[2(3)^{-1}=2/3\]

Answer 26

Ohhh I get it now ^_^ thanks

Answer 27

yw