Question

f(x)=sqrt((ln6x+7)(7x−3))

Answer 1

thank you for letting sam to have madel

Answer 2

f(x)=(X^2+9)^5/(x^2+6)^2

Answer 3

what is the question here?

Answer 4

logarithmic differentiation

Answer 5

which function?

Answer 6

@jim_thompson5910

Answer 7

@jim_thompson5910

Answer 8

@jim_thompson5910

Answer 9

can you post a screenshot please?

Answer 10

the problem is a bit ambiguous

Answer 11

@jim_thompson5910

Answer 12

http://www.mathcentre.ac.uk/students/topics/differentiation/by-logs/

Answer 13

Ok sure

Answer 14

wait, its nowhere close to sqrt((ln6x+7)(7x−3))

are you sure this is the right one?

Answer 15

yes this question and root question

Answer 16

well the second one was clearer than the first, so no need for a screenshot (in my opinion) for the second one

Answer 17

ok, sorry about that

Answer 18

ok so it's completely flipped around, thought so, but wasn't sure

Answer 19

you would use the chain rule twice, then the product rule once you hit the product

Answer 20

for the root one?

Answer 21

yes

Answer 22

I see! what part do i use chain rule?

Answer 23

for the ln( ... ) part and the sqrt( ... ) part

Answer 24

(6x+7)^1/2 (7x-3)^1/2?

Answer 25

first you would take the derivative of ln(x) to get 1/x

but x is some expression with a square root of a product

Answer 26

so it's really 1/(that expression)

but then you use the chain rule to multiply it by the derivative of that expression

Answer 27

then you would use the chain rule again to derive (6x+7)(7x-3)

Answer 28

which you would use the product rule at this point

Answer 29

so as you can see, it gets ugly fast

Answer 30

in(6x+7)(in7x-3)?

Answer 31

it's L n (lower case L though)

Answer 32

and that seems too simple to be the final answer

Answer 33

no the first step, haha

Answer 34

oh, no that's not correct either

Answer 35

oh :(

Answer 36

the first step is to derive

ln( sqrt( (6x+7)(7x-3) ) )

to get

1/( sqrt( (6x+7)(7x-3) ) ) * d/dx[ sqrt( (6x+7)(7x-3) ) ]

Answer 37

you would then take the derivative of sqrt( (6x+7)(7x-3) )

Answer 38

take?

Answer 39

(6+7)'7x-3)^1/2?

Answer 40

the derivative of sqrt( (6x+7)(7x-3) ) is _____

Answer 41

what you got isn't quite it

Answer 42

In(6)(7)?

Answer 43

no just focus on sqrt( (6x+7)(7x-3) ) for now

Answer 44

ok, sqrt is always confusing me

Answer 45

if y = sqrt(x)

then y ' = 1/(2*sqrt(x))

Answer 46

Oh! 1/ (6x+7)(7x-3) )

Answer 47

well, before i derive it. try to expanding (6x+7)(7x-3)=42x^2+31x-21
so, f(x) = ln{sqrt(42x^2+31x-21)}
now, we derive it by chain rule

a re-explain what is @jim_thompson5910 said before,
if given y = sqrt(f(x)) then y ' = f '/2sqrt(f(x))
and we knowed that the derivative of ln(u) = 1/u

thus, the derivative of y can be
y ' = {(42x^2+31x-21)'/2sqrt(42x^2+31x-21)}/(sqrt42x^2+31x-21)
y ' = (84x+31)/2(42x^2+31x-21)

Answer 48

look not enough good to read :)
i will figure out that :