Question

plss answer this
Use derivatives of limits
y=7-4x/5+2x

Answer 1

you have to find y'?

Answer 2

yeah

Answer 3

Use "derivative of a fraction", so, your y=f/g, then f=7-4x and g=5+2x.
Then y'=(f'g-fg')/g^2

Answer 4

Use quotient rule

Answer 5

its not chain rule method .. its derivatives of limits

Answer 6

(-70-4x)/25

Answer 7

how did u get that ?? plss show us solution so i can understand it

Answer 8

you sure destiny? For me it's -34/(5+2x)^2 D:

Answer 9

My answer is wrong, I used a quotinent rule

Answer 10

\[\frac{ 7-4(x+Deltax) }{ 5+2(x+Deltax) }\] this should be the simplified.. im confused

Answer 11

y=(7-4x)/(5+2x) or y=7-4x/5+2x ???

Answer 12

\[\frac{ 7-4x }{ 5+2x } the problem :\]

Answer 13

okay, that's different, it will start with
\[\lim \frac{ \frac{7-4(x+h)} {5+2(x+h)} - \frac {7-4x} {5+2x} }{ h } h\to 0\]

Answer 14

h=delta X

Answer 15

ok

Answer 16

Your result should be the same. The difference is that you're obtaining the derivative by the limit's definition of it.

Answer 17

why dont you use quotient rule much more easier

Answer 18

is chain rule and derivatives of limits is the same ??

Answer 19

it is not chain rule it is quotient rule lol ya it is the same and faster

Answer 20

ok

Answer 21

All the derivative rules were obtained from the definition, so, yes, if you can use 'em, use 'em xD

Answer 22

unless you are required to use the classic way lol

Answer 23

tnx for all the help