Question

find the derivative of each of the following

Answer 1

\[f(x)=\frac{ 6x-11}{ 8x+1 }\]

Answer 2

@ganeshie8 what would we do in this situation

Answer 3

familiar wid quotient rule ?
or product rule ?

Answer 4

not really

Answer 5

you need to knw any one of them to do this

Answer 6

\(\large \left(\frac{f}{g}\right)' = \frac{g\times f' - g' \times f}{g^2}\)

Answer 7

i am confused, for f or g whay would we plug in

Answer 8

\(\large \left(\frac{ 6x-11}{ 8x+1 }\right)' = \frac{(8x+1)\times (6x-11)' - (8x+1)'\times (6x-11)}{(8x+1)^2}\)

Answer 9

then we multpliy the equations and subtract them

Answer 10

take the derivatives first

Answer 11

\(\large \left(\frac{ 6x-11}{ 8x+1 }\right)' = \frac{(8x+1)\times (6x-11)' - (8x+1)'\times (6x-11)}{(8x+1)^2}\)


\(\large \left(\frac{ 6x-11}{ 8x+1 }\right)' = \frac{(8x+1)\times (6) - (8)\times (6x-11)}{(8x+1)^2}\)

Answer 12

now multiply and simplify

Answer 13

\[\frac{ -82 }{ 80x+1 }\]

Answer 14

the derivative is 80x which is 80

Answer 15

?? is this another problem ?

Answer 16

no isnt that the answer

Answer 17

nope

Answer 18

\(\large \left(\frac{ 6x-11}{ 8x+1 }\right)' = \frac{(8x+1)\times (6x-11)' - (8x+1)'\times (6x-11)}{(8x+1)^2}\)


\(\large \qquad \qquad ~~= \frac{(8x+1)\times (6) - (8)\times (6x-11)}{(8x+1)^2}\)


\(\large \qquad \qquad ~~= \frac{48x+6 - 48x +88}{(8x+1)^2}\)

Answer 19

simplify

Answer 20

@ganeshie8 it would become \[\frac{ 94 }{ (8x-1)^2 }\]

Answer 21

how come denominator changed to 8x-1 ha ?

Answer 22

it should be (8x+1)^2 in the denominator right ?

Answer 23

@ganeshie8 my mistake, but that would correct though\[\frac{ 94 }{ 8x+1 }^2\]

Answer 24

power of two is for denominator

Answer 25

\(\large \frac{94}{(8x+1)^2}\) is \(\large \color{Red}{\checkmark}\)

Answer 26

but would the derivative be 8

Answer 27

derivative is \(\large \frac{94}{(8x+1)^2} \)