Question

f(x)=sqrt(x+7). a=9. Find the derivative.

Answer 1

I plugged into into the formula f(x)-f(a) divided by x-a. Had to multiple by the conjugate, and i simplified it down to. 1/((sqrt x+7)+4)

Answer 2

Do i then plug in 9 for x? Which would give me 1/8. Not sure if that's the right way to approach the problem though.

Answer 3

can u post the ques in a better way?

Answer 4

Sure thing, screenshot attached.

Answer 5

\[\Huge f'(x)=\frac{1}{2 \sqrt{x+7}}\]
Question now asks for (a,f(a))
=>(9,4)
since f(a)=sqrt{9+7}

Answer 6

Oh i used the wrong method for the derivative i guess. I'll try it the other way and see if i get that.

Answer 7

Yeah for part b, it wants it in the form y-4=m(x-9) i think. Sound about right? After i plug in slope, and solve for y?

Answer 8

Also is the derivative the simplest form of the slope? Normally its mx, but there is an x already in the derivative so it would be 1/(2(sqrt(x+7))x ?

Answer 9

slope=dy/dx=f'(x)

Answer 10

Correct the derivative, which i solved for.

Answer 11

u wont have a x there

Answer 12

@tcarroll010 Thanks thats exactly what i did, the first time i just used the wrong formula. Now i'm trying to solve for y the tangent line equation.

Answer 13

that is what i said :D

Answer 14

\[y-4=2\sqrt{x+7}(x-9) \]

Answer 15

Then i'd simplify?

Answer 16

Ohh my bad, should be 1/2(sqrt(x+7))

Answer 17

Isn't m = to the entire f'(x)?

Answer 18

\[y-4=(\frac{ 1 }{ 2\sqrt{x+7} })(x-9)\]

Answer 19

Is that correct, before i simplify and solve for y?

Answer 20

Okay this is embarrassing, but how do i simplify that algebraically for y? The double x and the square root confused me.

Answer 21

\[y=(\frac{ 1 }{ 2\sqrt{x+7} })x-9(\frac{ 1 }{2\sqrt{x+7}}) +4 \]

Answer 22

Right this is what i have, but can that simplify, normally we have to have it simplified, but those radicals are a mess :/

Answer 23

But if's supposed to factor through to both the x and the 9 i thought?

Answer 24

\[y=(\frac{ x-9 }{ 2\sqrt{x+7} }) +4\]

Answer 25

So just this is correct then?

Answer 26

THanks : )

Answer 27

Unfortunately thats wrong, i can try putting the 4 in the numerator, but normally it will say your answer is correct but not simplified. Is there something else wrong? Opening this up for someone else to help. Screenshot of problem, and answer error.

Answer 28

I guess it could be this?

\[y=\frac{ x-9+8\sqrt{x+7} }{ 2\sqrt{x+7} } \]

Answer 29

URG. Turns out the answer was...

\[y=\frac{ 1 }{8 }x +\frac{ 23 }{ 8 }\]

No clue, i knew that algebra was too messy to make sense. Well when you're back hopefully you can help @tcarroll010

Answer 30

For the line tangent to our curve at x=a, we want it in slope-intercept form:\[\large y=mx+b\]

You already found \(\large m\), the slope of the tangent line by calculating the derivative at \(\large a\).
\(\large m=f'(a)\)

\[\large y=\frac{1}{8}x+b\]

You did so using the "other" limit definition of the derivative.
\[\large \lim_{x \rightarrow a}\dfrac{f(x)-f(a)}{x-a}\]

To solve for \(\large b\), the y-intercept, you need to plug in a point that lies on the line.
The point \(\large \left(a,f(a)\right)\) will lie on the line.

Plugging \(\large a\) into the original function gives us,\[\large f(9)=\sqrt{9+7}=16\]So we've come up with the coordinate pair, \(\large (9,16)\), which we can plug in for x and y in our equation to solve for b!

Understand how that works?
\(\large y=\dfrac{1}{8}x+b \qquad\rightarrow\qquad 16=\dfrac{1}{8}\cdot9+b\)

Answer 31

Woops I miscalculated \(\large f(9)\) lol
Should be 4 I guess :)

Oh well, hopefully that helps some.

Answer 32

Oh i see. So we don't actually plug the derivative of f(x) in as slope. You have to plug a into f'(x) to solve for 1/8 instead of the radical. Thats where it went wrong, everything else was the same.

Thanks it makes perfect sense now. : )

Answer 33

cool :)