Question

y=f(x)=8/sqrt(x-2), (6,4)...differentiate the functions then find an equation of the tangent line at the indicated point...f(x)=f(x+h)-f(x)/h is this the formula i am supposed to use????

Answer 1

good god you have to do it by hand?

Answer 2

yes. i knw the end answer is supposed to look like y-y1=m(x-x1) but im having trouble finding the slope

Answer 3

do we have to use the differnce quotient to take the derivative?

Answer 4

\[f'(6)=\lim_{h\to 0}\frac{f(6+h)-f(6)}{h}\]

Answer 5

\[=\frac{\frac{8}{\sqrt{6+h-2}}-4}{h}\]

Answer 6

\[=\frac{\frac{8}{\sqrt{4+h}}-4}{h}\]

Answer 7

now subtract and get
\[=\frac{\frac{8-4\sqrt{4+h}}{\sqrt{4+h}}}{h}\]

Answer 8

so would the sqrt (4+h) cancel out?

Answer 9

oh hell no

Answer 10

we have to multiply top and bottom by the conjugate to get rid of the radical in the numerator

Answer 11

ohhhhhhhhhhhhh kk

Answer 12

\[=\frac{\frac{8-4\sqrt{4+h}}{\sqrt{4+h}}}{h}\]
\[=\frac{8-4\sqrt{4+h}}{h\sqrt{4+h}}\]

Answer 13

you can factor out an 4 now if you like or just multiply as follows
\[=\frac{8-4\sqrt{4+h}}{h\sqrt{4+h}}\times \frac{8+4\sqrt{4+h}}{8+4\sqrt{4+h}}\]

Answer 14

jesus this is a pain, it is much easier to take the derivative using rules, but ok whatever

Answer 15

\[\frac{64-16(4+h)}{h(\sqrt{4+h})(8+\sqrt{4+h})}\]

Answer 16

end up with
\[\frac{-16h}{h(\sqrt{4+h})(8+\sqrt{4+h})}\]

Answer 17

FINALLY you can cancel the h, which was the goal all along and get
\[\frac{-16}{(\sqrt{4+h})(8+\sqrt{4+h})}\]

Answer 18

and now take the limit by replacing h by zero. get
\[\frac{-16}{(\sqrt{4})(8+\sqrt{4})}\]

Answer 19

or
\[-\frac{4}{5}\]

Answer 20

ok s73 u have no idea how appreciative i am of all this, but the book has -4/(x-2)sqrt(x-2) as an answer...

Answer 21

and they have -1/2 as a slope..=/

Answer 22

maybe i made a mistake, let me check

Answer 23

yes it is -1/2 damn

Answer 24

oh because i dropped a damn 4 somewhere!

Answer 25

haha feel free to use the shortcut u were talking bout earlier

Answer 26

\[=\frac{8-4\sqrt{4+h}}{h\sqrt{4+h}}\times \frac{8+4\sqrt{4+h}}{8+4\sqrt{4+h}}\]

Answer 27

\[\frac{64-16(4+h)}{h(\sqrt{4+h})(8+4\sqrt{4+h})}\]

Answer 28

\[\frac{-16}{(\sqrt{4+h})(8+4\sqrt{4+h})}\]

Answer 29

there, now replace h by zero!

Answer 30

f(x)=8/sqrt(x-2) = 8 * (x-2) ^(-1/2)
f'(x) = 8 * (-1/2) * ( x -2) ^(-3/2)
= -4/ (x -2) ^(3/2)

At ( 6, 4):
f'(6) = -4/ ( 6 -2) ^(3/2) = -1/2
Tangent line:
y -4 = (-1/2) (x - 6)
->y = -x/2 +3 + 4
y = -x/2 + 7

Answer 31

\[\frac{-16}{(\sqrt{4})(8+4\sqrt{4})}\]

Answer 32

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

Answer 33

lot easier to use the short cut rules for finding the derivative, then it is to use the difference quotient. but it is still good practice.
i dropped a 4 when i was calculating. i put it back and hope the steps are clear

Answer 34

for the bottom i'm gettin (2)(8+4(2)) is that correct?

Answer 35

yes

Answer 36

numerator is -16, denominator is 2(8+8)=32, get -1/2

Answer 37

wow haha well thank u for this lesson haha, man i need to learn the rules for deriv. i hope my professor teaches em soon. THANK U S73!!

Answer 38

i owe u at least a beer dude

Answer 39

i'll take a nice porter, maybe anchor porter. thanks