Question

f'(x) where f(x) = ln((ex^2)/(x-5)^7)

Answer 1

\[f(x) = \ln(\frac{ ex^2 }{ (x-5)^7 })\]

Answer 2

Use properties of log first

Answer 3

That will make the differentiation process tons easier

Answer 4

so\[\frac{ lnex^2 }{ \ln (x-5)^7 }\]

Answer 5

That isn't a property of log

Answer 6

Can I have a hint?

Answer 7

\[ex^2 \ln (x-5)^7\]

Answer 8

My hint is that you review the properties of log

Answer 9

ln(a/b) =ln(a) - ln(b)

Answer 10

\[\large \rm \ln\left(\dfrac{\spadesuit }{\heartsuit }\right) = \ln\left(\spadesuit\right)-\ln\left(\heartsuit\right)\]

Answer 11

\[\large \rm \ln\left(\spadesuit \heartsuit \right) = \ln\left(\spadesuit\right)+\ln\left(\heartsuit\right)\]

Answer 12

http://www.purplemath.com/modules/logrules.htm

Answer 13

okay so ln (ex^2)- ln (x-5)^7

Answer 14

you can work this without logs if u dont like them, bit it would be very messy using quitient and product rules

Answer 15

You can even apply the power rule for log

Answer 16

lmao, spades and hearts huh?

Answer 17

2ln ex - 7 ln (x-5)

Answer 18

How far off am I

Answer 19

Good job

Answer 20

Wait was that 2 on the ex or the x

Answer 21

its \[ex^2\]

Answer 22

So the second term you have is right but the first term is a bit off

Answer 23

lol i got those from satellite i think...
hey \(\large \ln (ex^2) = \ln(e) + \ln(x^2)\)

Answer 24

if it's \(ex^2\) then e is a function and so is \(x^2\), therefore product rule is applied.

Answer 25

Oh log rule of the numerator alone. It's like inception.

Answer 26

haha so ln(e) + ln (x^2) - 7 ln (x -5)

Answer 27

Seems better. Does everyone else agree?

Answer 28

\[\frac{ 1 }{ e} + \frac{ 1 }{ x^2 } - 7 \frac{ 1 }{ x-5 }\]

Answer 29

do I need chain rule?

Answer 30

Yea we can do one more thing but I think this should be easy as is already to differentiate

Answer 31

you're partially right there

Answer 32

With log functions, you have a function inside of a function, you most always use chain rule

Answer 33

ln(e) is a constant

Answer 34

There's exceptions sometimes.

Answer 35

so if ln(e) is a constant that is = 0

Answer 36

Remember, \(\ln (e) = \log_{e} e = 1\)

Answer 37

And you need to look at your second term also

Answer 38

\[\frac{ 1 }{ x^2} * 2x - 7 \frac{ 1 }{ x - 5 } * 1\]

Answer 39

\[\frac{ 1 }{ e} + \frac{ 1 }{ x^2 } - 7 \frac{ 1 }{ x-5 }\] first part: constant - deirvative = 0
second part: you still got an x^2 - take its derivative,
3rd part - x-5 is sill a function, and the derivative of x =1, so what are you going to do now

Answer 40

Is it what I posted above?

Answer 41

\[\frac{ 1 }{ e} + \frac{ 1 }{ x^2 } - 7 \frac{ 1 }{ x-5 }\]\[0 + \frac{1}{x^2} \cdot 2x -\frac{7}{x-5}\cdot 1\]\[\frac{2x}{x^2} - \frac{7}{x-5}\]\[2x^{1-2} -\frac{7}{x-5}\]\[2^{-1} -\frac{7}{x-5}\]\[\boxed{\large \frac{2}{x} -\frac{7}{x-5}}\]There you have it

Answer 42

You were almost there. You have any questions about it?

Answer 43

not sure if you have to totally simplify it either with the common denominator and stuff, that would make it look pretty but it'd hurt.

Answer 44

I did \[2x/(x^2)-7/(x-5)\]

Answer 45

Well, you don't want to leave it like that because the x's inn the first fraction can indeed simplify further.

What I meant was, once you have your boxxed answer, you can actually simplify it further by obtaining the LCD of the two functions, and this and that.

Answer 46

Okay, well thanks for your assistance.

Answer 47

No problem. IF you are interested in simplifying it further, it would be like this:\[\frac{2}{x} -\frac{7}{x-5}\]\[\frac{2x-10}{x^2 -5x} -\frac{7x}{x^2-5x}\]\[\boxed{\large \frac{-9x-10}{x^2-5x}}\]