Question

Equation of a tangent to y =

Answer 1

\[\huge y=\frac{ 18 }{ (x+2)^2 }\] at point (1,2)

Answer 2

Do you recall how the derivative is related to the tangent ?

Answer 3

i am not allowed to derive, this is for an advanced function's class

Answer 4

Are you allowed to take limits ?

Answer 5

yes

Answer 6

\[\large \lim_{h \rightarrow 0}\frac{ \frac{ 18 }{ ((x+h)+2)^2 }-\frac{ 18 }{ (x+2)^2 } }{ h }\] like this ? @logicalapple

Answer 7

That is exactly right. What do you obtain when evaluating this limit ?

Answer 8

just curious why you are allowed to take limits, but not derivatives, which themselves are limits

Answer 9

expanding this is confusing

Answer 10

@satellite73 because it's an advanced functions class and according to our teacher we can only do limits

Answer 11

satellite73 is right, it's actually harder to solve this problem taking limits instead of derivatives...

Answer 12

forget about the \(h\) in the denominator, and the common factor of \(18\) which you can put in at the end, and concentrate on
\[\frac{1}{(x+h)^2+2}-\frac{1}{(x+h)^2}\]

Answer 13

damn typo

Answer 14

\[ \frac{ 1}{ ((x+h)+2)^2 }-\frac{ 1 }{ (x+2)^2 } \]

Answer 15

you get
\[\frac{(x+2)^2-(x+h+2)^2}{(x+h+2)^2(x+2)^2}\]

Answer 16

the numerator is 18

Answer 17

i know , i ignored it to make life easier
there is a common factor of 18 so you can take the limit and the multiply the result by 18

Answer 18

ohh okay:P

Answer 19

multiply all this crap out, everything without an \(h\) in it will be gone, then you can divide by \(h\)

Answer 20

i mean multiply out in the numerator, leave the denominator in factored form. don't touch it

Answer 21

I'll do it taking derivatives just for you to appreciate the difference

Answer 22

I mean, i don't understand professors sometimes

Answer 23

if you do it carefully, your numerator will be
\[-2xh-4h-h^2\] then when you divide by \(h\) , the one i ignored in the denominator, you get
\[-2x-4-h\]

Answer 24

now you should be looking at
\[\frac{-2x-4-h}{(x+2)^2(x+h+2)^2}\] and take the limit by simply replacing \(h\) by 0

Answer 25

oh lord i am an idiot, it is much easier than this!!

Answer 26

\[y'=\frac{ -36 }{ (x+2)^3 }, Slope(m) = y'(1) ->\frac{ -36 }{(1+2)^3 }= -4\]

Answer 27

You already got the points and the slope, so apply point slope ecuation \[y-y1=m(x-x1) -> y-2=-4(x-1)\]

Answer 28

and it's done

Answer 29

what satellite is doing is applying the definition of the derivative, which is much harder that what i've done. Once you learn derivatives it'll become easier.

Answer 30

Rule of the four steps.

Answer 31

i'll get this right eventually
use
\[\lim_{x\to 1}\frac{\frac{18}{(x+2)^2}-2}{x-1}\] that is what i wanted

Answer 32

@GCR92 i am not allowed to use derivatives

Answer 33

then use the second one i wrote, it is easier than the first one

Answer 34

the algebra still sucks though

Answer 35

hmm okay

Answer 36

ignoring the denominator again you get
\[\frac{18}{(x+2)^2}-2=\frac{18-2(x+2)^2}{(x+2)^2}\]
\[=\frac{18-2x^2-8x-8}{(x+2)^2}\]
\[=\frac{-2x^2-8x+10}{(x+2)^2}\]
\[\frac{-2(x-1)(x+5)}{(x+2)^2}\]

Answer 37

now divide by \(x-1\) and get
\[\frac{-2(x+5)}{(x+2)^2}\] and finally you can replace \(x\) by \(1\) aka take the limit as \(x\to 1\) and get your slope

Answer 38

what a colossal pain, but that will work