Question

Differentiate the function. Then find an equation of the tangent line at the indicated point on the graph of the function.

Answer 1

\[f(x)=6+\sqrt{1-x}\] at (-8,9)

Answer 2

do you know that you can write the square root as an exponent to the 1/2 power?

\[ f(x)= 6 + (1-x)^{\frac{1}{2} }\]

you can use the "power rule" to take the derivative.

Answer 3

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

Answer 4

you lost me

Answer 5

do you know the derivative of
\[ x^2\] ?

Answer 6

2

Answer 7

what is the derivative of 2x ?

Answer 8

2...

Answer 9

yes the derivative of 2x is 2

the derivative of x^2 is 2x

if you have time, watch this
http://www.khanacademy.org/math/calculus/differential-calculus/power_rule_tutorial/v/power-rule

it will be easier to explain after you watch

Answer 10

I understand simple derivatives like that. I just don't know what to do with this problem

Answer 11

what is the derivative of x^3 ?

Answer 12

3x^2

Answer 13

what is the derivative of
\[ x^{\frac{1}{2} } \]?

Answer 14

1/2x?

Answer 15

same rule: multiply by the 1/2 and change the exponent by subtracting 1

1/2 -1 is the new exponent. try again.

Answer 16

\[\frac{ 1 }{ 2 }x ^{-\frac{ 1 }{ 2 }}\]

Answer 17

next, what is the derivative of
(x+1)^2 ?

Answer 18

2x?

Answer 19

we use the "chain rule" to take the derivative of (x+1)^2

the first step is let the problem be u^2 (where u is x+1)
the derivative of u^2 is 2u
u is x+1, so 2u is 2(x+1)

2nd step: after getting 2u we then multiply by du (the derivative of u)
u is x+1 and du is d(x+1) = d(x) + d(1)= 1+0= 1

all that work to get du=1.

we find d(x+1)^2 is 2(x+1)

Answer 20

okay, then what?

Answer 21

if you followed that, you can do your problem

Answer 22

\[f ^{1}(x)=2(x+1)\]

Answer 23

try doing
\[ f(x)= 6 + (1-x)^{\frac{1}{2} } \]

the derivative of the constant 6 is 0
you can try the rest...

Answer 24

\[f ^{1}(x)=\frac{ 1 }{ 2 }x ^{-\frac{ 1 }{ 2 }}\]

Answer 25

except that is the answer of the derivative of
\[ f(x)= 6 + x^{\frac{1}{2} } \]

you must keep the (1-x)
and then multiply by the derivative of (1-x)

Answer 26

I give up...

Answer 27

watch this for examples...
http://www.khanacademy.org/math/calculus/differential-calculus/chain_rule/v/chain-rule-examples

Answer 28

okay, last try. \[f ^{1}(x)=\frac{ 1 }{ 2 }(1-x)^{-\frac{ 1 }{ 2 }}\]

Answer 29

close. the last step is multiply that by the derivative of (1-x)

Answer 30

what do you mean?

Answer 31

first, what is d (1-x) ?

Answer 32

-1?

Answer 33

yes

what you are doing is following this rule (see the videos on why)
\[ d u^n = n u^{n-1} du \]

in your problem u is (1-x) and n = 1/2 and du is d (1-x) = -1
put it all together:
\[ f '(x)=-\frac{ 1 }{ 2 }(1-x)^{-\frac{ 1 }{ 2 }} \]

Answer 34

btw, it's f prime (short for df/dx) not f^1

Answer 35

okay thank you. now how do I find the equation of a tangent line?

Answer 36

the derivative of a function represents the slope of the tangent line when you evaluate at a particular x. in your case, f'(x) evaluated at x= -8 is the slope of the tangent line at x=-8

after you find the slope, you have the slope and a point (-8,9), so you can find the equation of a line using that info.

Answer 37

what's the slope?

Answer 38

slope is what you learn about before you take calculus.

Answer 39

I know what a slope is....I just don't know how to find it in these problems. I'm taking this online, am 16 and doing my best. thanks, not.

Answer 40

in that case, you might want to watch some of Khan's videos.

I posted above how to find the slope:
f'(x) evaluated at x= -8 is the slope of the tangent line at x=-8

that means replace x with -8 in your derivative, do the arithmetic to simplify to a number.
that number is the slope of the tangent line.

Answer 41

m= -1/6

Answer 42

and the tangent line is \[y=-\frac{ 1 }{ 6 }x+\frac{ 23 }{ 3 }\]

Answer 43

yes, looks good

Answer 44

Thank you!