Question

Need help with this derivative problem!
d/dx ln(4x-x^2)

Answer 1

\[\frac{d}{dx}[\ln(f(x)]=\frac{f'(x)}{f(x)}\]

Answer 2

wait so I take the derivative and put it over the original problem?

Answer 3

oh wait nevermind I think I get it. Haha. Just one minute I am going to try and solve it.

Answer 4

4-2x/4x-x^2????

Answer 5

@satellite73 ???????

Answer 6

Correct

Answer 7

Awesome! I just didn't know the formula. :)

Answer 8

Derivative of In(x) is 1/x

Answer 9

try the chain rule. We know that the derivative of ln(x) is 1/x so we should write 1/(4x-x^2)

then we will need to multiply it by the derivative of the inside function (the 4x-x^2) so do that.

If you put it all together you will see how it resembles satellite's formula!

Answer 10

Is the formula the same for this one too??
ln(x(sqrt(x+1)???

Answer 11

well the derivative of that problem.

Answer 12

Yes except there will be 3 steps for the chain rule. First take care of the ln, then the square root (which you can change to a fraction exponent) and then the inside function.

Answer 13

I tried doing the same thing but I got 3x+2/2(sqrt(x+1)/x(sqrt(x+1) and I do not believe that is the correct answer.

Answer 14

You just need to remember that the derivative of ln(x) = 1/x

Answer 15

I will try it out

Answer 16

Whoops I gave you some wrong info. You will actually need to use the product rule on the inside of that function!

Answer 17

Yeah I am not quite sure how you got to your answer

Answer 18

Do you want to go through it step by step?

Answer 19

haha. I am not sure how to get to the right answer. Yes that would be so helpful!

Answer 20

Okay so first we see there is an ln on the outside, so lets take the derivative of that to get 1/(xsqrt(x+1))

Answer 21

The answer is 1/x + 1/2(sqrt(x+1)^2

Answer 22

Could someone please verify if it is right?

Answer 23

Now we want to take the derivative of the stuff that is INSIDE the ln function. We can see that we will need to use the product rule. Have you familiar with it?

Answer 24

inside okay let me see if I can solve it....

Answer 25

the (x+1) part?

Answer 26

think of the square root of (x+1) as (x+1)^(1/2)

Answer 27

so we have x(x+1)^(1/2) and we want to find the derivative of these two things

Answer 28

we can't exactly multiply that x into the (x+1)^(1/2), so we must use the product rule which is d/dx f(x)g(x)=f(x)'g(x)+f(x)g(x)'

Answer 29

I got 3x+2/s(x+1)^1/2

Answer 30

Is that your full answer?

Answer 31

3x+2/(2(x+1)^1/2

Answer 32

no just for that inside part

Answer 33

ok I will try it

Answer 34

okay I think you have made a mistake somewhere in there. Did you use the product rule?

Answer 35

I don't know what rule I used. I used the one that I know.

Answer 36

the f(x)g(x)=f(x)'g(x)+f(x)g(x)' one?

Answer 37

that is meant to say the derivative of f(x)g(x)

Answer 38

Does that look familiar? You make know it as 'take the derivative of the first times the second... plus the derivative of the second times the first'

Answer 39

oh ya I know that one

Answer 40

and do you know how to take the derivative when there are exponents involved?

Answer 41

not really

Answer 42

when taking the derivative of the (x+1)^(1/2) move the exponent (1/2) over to the front of the (x+1) and subtract 1 from the exponent. This should probably seem really familar to you.

The only difference is that now we have more stuff under the exponent (we have an x+1) so we must multiply what we get by the DERIVATIVE of the x+1