Question

f(x)=sqrt(x+3); find f'(1)

Answer 1

find f'(x) and then plug 1 into it to get the answer

Answer 2

im having trouble understanding how to find f'(x) though
i dont understand the square root part of it

Answer 3

x=2

Answer 4

find f(1) is basically asking you to plug a "1" in everywhere you see a "x"

Answer 5

f(1)= sqrt(1+3)
f(1)=sqrt(4)
f(1)=2

Answer 6

f(1) = 2 or f'(1) = 2 ??, im lost

Answer 7

f'

Answer 8

i didn't see that ' there sorry :X

Answer 9

i dont have 2 as an option for an answer

Answer 10

hmm that's weird...what are the answers ?

Answer 11

1/4, sqrt7/4, 5sqrt7/4, 5/4

Answer 12

O__O

Answer 13

sorry Meepi, I post it before seeing you are working on.

Answer 14

f(x)=sqrt(x+3); find f'(1), can you help me with this one please. the answer appears to me to be 1/2, but it's not

Answer 15

f' is the derivative,
\(\Large f(x) = \sqrt{x +3}=(x + 3)^{{1 \over 2}}\)

Use the chainrule + power rule to compute the derivative:
\(\Large f'(x) = \frac{1}{2\sqrt{x+3}}*1=\frac{1}{2\sqrt{x+3}}\)

\(\Large f'(1) = \frac{1}{2\sqrt{1 + 3}} = \frac{1}{3}\)

Answer 16

whoops that \(\frac{1}{3}\) should be \(\frac{1}{4}\)

Answer 17

okay 1/4 is an option so i'll take it

Answer 18

still there?

Answer 19

Yeah

Answer 20

f(x)=sqrt(x+3); find f'(1), can you help me with this one please. the answer appears to me to be 1/2, but it's not

Answer 21

Do you know how to take a derivative?

Answer 22

in some cases... i know this is basic, but im lost

Answer 23

i dont get it when its root or fractions

Answer 24

Did you learn the chain rule?

Answer 25

nope

Answer 26

We can start easy, do you know the derivate of \(\sqrt{x}\)?

You can use the power rule to calculate it if you write it as \(\Large x^{1 \over 2}\)

Answer 27

yes i get that

Answer 28

Alright so you should get:
\(\Large \frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2}\frac{1}{\sqrt x}=\frac{1}{2\sqrt{x}}\)

Now basically what the chain rule means is that if you have a function say
\(f(x) = \sqrt{x + 3}\)

You can take the derivative by pretending that what's under the root is just a single variable, as long as you multiply the result by the derivative of what it actually is.

so lets just pretend x + 3 is one variable u, then according to the chain rule we can differentiate it as long as we multiply it by the derivative of u:
\(f(x) = \sqrt{x + 3}\)
Pretend x+3 is one variable say u:
\(f(u) = \sqrt{u}, u = x + 3\)
take the derivative normally, but don't forget you have to multiply by the derivative of u because you just pretend it's a single variable (and it actually isn't)
\(\Large f'(x) = \frac{1}{2\sqrt u}\cdot u'\)
The derivative of x + 3 is just 1 so:
\(\Large f'(x) = \frac{1}{2\sqrt u}\)
Now just fill in u again:
\(\Large f'(x) = \frac{1}{2\sqrt{x + 3}}\)

\(\Large f'(1) = \frac{1}{2\sqrt{1+3}} = \frac{1}{2\sqrt{4}} = \frac{1}{2*2}=\frac{1}{4}\)

Answer 29

so what do i do in the case of a fraction like f(x)= -1/x+6 finding f'(4)?

Answer 30

You use the quotient rule:
\(\Large f(x) = \frac{p(x)}{q(x)}\)
\(\huge f'(x) = \frac{p'(x)q(x) - p(x)q'(x)}{\left(q(x)\right)^2}\)