Question

Find the equation of the tangent line to the curve y=√x at x=4.

please explain step by step? thank you!! :)

Answer 1

the derivative at the point (4,2) is the slope of tangent line.

then use point-slope form to get equation of line
y-y1 =m(x-x1)

Answer 2

so y-2 = m(x-4) ?

and how would we find m ?

Answer 3

slope = derivative

Answer 4

okay :)

would we use f(a+h)-f(a)
---------
h

??

Answer 5

you could...have you learned the power rule for finding derivatives of polynomials ?

Answer 6

yes :)

f(x)=x^n
f ' (x) = nx^(n-1)

right? :)

Answer 7

\[\frac{d}{dx} x^n = n x^{n-1}\]

Answer 8

haha yeah use that

Answer 9

\[\Large y = √x = x^{1/2}\]

Answer 10

haha okie :P

so we would have

y=√x = x^(1/2) = (1/2) * x^((1/2)-(-.5) ?

Answer 11

^idk what you're doing with the exponent. Reduce it by 1 (subtract 1 from it)

Answer 12

oh oops :p

okay so y=√x = x^(1/2) = (1/2) * x^((1/2)-1)
= (1/2) * x^-1/2 ?

=(1/2)x
-------
x^(1/2)

??

Answer 13

Why's there an extra x...

Answer 14

oh oops typo haha

so = x/x^(1/2) ?

Answer 15

Still an extra x.

This is right: = (1/2) * x^-1/2
so\[\Large y ' = \frac{ 1 }{2 \sqrt x }\]

Answer 16

ohh okay... so that's the final equation?

is there any way you can write out what happens from the first step to the last step?

i'm kinda confused on where they connect :(

Answer 17

Given f(x) and point (x1,y1)

equation of tangent line is:
\[y - y_1 = f'(x_1) (x - x_1)\]

Answer 18

ohh okay..

so then the points are (4,2)

and then we get

y=√x = x^(1/2) = (1/2) * x^((1/2)-1)
= (1/2) * x^-1/2
= y ' = 1/ (2√x)

? and that is the final equation of the tangent line to the curve y=√x at x=4?


did i miss anything? :/

Answer 19

That is not the tangent line. That y' gives you the slope, m, in your tangent line: y = mx + b

Answer 20

Plug in x=4 to find the slope.

Answer 21

ohh okay

but those steps above are correct so far?

Answer 22

okay, so 2=(1/(2√x)) * (4) + b

?

Answer 23

Find m FIRST using y ' = 1/ (2√x) and x=4

Answer 24

ohh i see..

so y' = 1/ 4 ?

Answer 25

oh and is all this using the definition of the slope of the tangent line?

i'm supposed to do learn it via definition right now, not shortcut :/ just wondering hahaa

Answer 26

Idk if this is really a shortcut.

Now you have m, in y = mx+b

and you have the point (4,2). Find b.

Answer 27

haha okie :P

so 2=(1/4)(4) + b
= 2=1 +b
= b=1 ?

Answer 28

Yep, now write the line y=mx+b

Answer 29

\[y=\frac{ 1 }{ 4 }x + 1 \]

?? is that right?

Answer 30

Yep

Answer 31

awesome! so that's the final equation of the tangent line? ?:O

Answer 32

Yep. It's the same process every time.

Find the slope at the point (using the derivative).

Find the line y=mx+b using that slope, and the point it passes through.

Answer 33

ahh okay awesome!!

and also wondering, would you be able to solve this with the

f(a+h) - f(a)
----------
h

equation??

Answer 34

or is this the only way to solve for the equation?

Answer 35

That just is the long way of finding the slope, m.

Answer 36

ohh okay... just for future reference, how would i find the slope the long way?

Answer 37

would i be plugging in like this?
\[\lim_{h \rightarrow 0} \frac{ (f(4+h)-f(4) }{ h }\]

??? ;/ @agent0smith

Answer 38

Yeah, I think that's right.

Answer 39

okay:)

what would happen next? ;/

Answer 40

You'd simplify it until you have the slope, 1/4

Answer 41

okay cool, thank you!!!