Question

what is the derivative of cos(2x)/logx

Answer 1

by log x do you mean natural log (lnx) or base 10?

Answer 2

natural log

Answer 3

nooo base

Answer 4

I meant base

Answer 5

Well, when no base is written its an invisible 10. But a lot of people write logx to mean lnx. Ijust want to make sure which it is.

Answer 6

yeah its the one with 10

Answer 7

Okay, cool. So we have to do quotient rule then

\[\frac{ f'(x)g(x) - f(x)g'(x) }{ [g(x)]^{2} }\] Now the derivatives of the logs that are not ln work like this:

\[\frac{ d }{ dx }\log _{a}u = \frac{ 1 }{ (lna)u }*\frac{ du }{ dx }\]

Okay, so first we have derivative of cos(2x) which is -2sin2x, times our logx. Then minus cos(2x) timesderivative of logx, which would be 1/x(ln10). All of this is divided by (logx)^2

\[\frac{ -2\sin(2x)logx - \frac{ \cos(2x) }{ xln(10) } }{ (logx)^{2} }\]

How much do you think you need to have it simplified?

Answer 8

I got that :)) :)) I couldn't figure out what to do with the fraction on the top

Answer 9

yes simplified please

Answer 10

Well, I guess we can play with it and see what happens.

Answer 11

\[\frac{ -2xsin(2x)\ln(10)-\cos(2x) }{ xln(10)(logx)^{2} }\]

Im not even sure if theres nything worth doing to that.

Answer 12

Thank you... for trying

Answer 13

can I do another problem and show you if did it right?

Answer 14

Well, theres honestly nothing that I can imagine even trying. Unless youre sure it can be dumbed down, it just doesnt looklike it. And yeah, go for it.

Answer 15

yay :D
ok the problem is find y' of sin(5x)cos(5x)
I got an answer which is (cos5x)5cos(5x) + 5sin(5x)-5sinx(5x)

Answer 16

I have a feeling it should be simplified but it seems like the simplifying part is harder than finding the derivatives

Answer 17

Its supposed to beharder. Thats not just you. Calculus is often easy, its the algebra after that sucks xDD But alright, lets see

\[5\cos^{2}(5x) -5\sin^{2}(5x)= 5[\cos^{2}(5x)-\sin^{2}(5x)] = 5\cos(10x)\]

Answer 18

I know right! And I see what you did thanks. :D

Answer 19

Alright, cool xD Yeah, that one actually looked like it could be simplified, the otherone no xD

Answer 20

Lol true! simplifying and deciding whether to use product rule and chain rule kills me.

Answer 21

well, think of it this way. You always use chain rule, its just a lot of the time the chain rule leads to multiplication by 1, which is useless.

Answer 22

smh yeah it just confuses me sometimes, because I start with chain rule and then i have to do product rule at some point

Answer 23

If one of the layers happens to bea product rule then yeah.

Answer 24

But I think I got it now I just need to learn how to simplify my work

Answer 25

alrighty, cool :3

Answer 26

you got time for more problems?

Answer 27

yeah, ill see what I can do.

Answer 28

cool how do you find y' of 3sin(8x)cos(8x)

Answer 29

Its like the same problem. I betcha the answer is 24 cos(16x), haha.

Answer 30

LOL we'll see

Answer 31

Try to work it out and see xD

Answer 32

I'm used to having the exponent so i can multiply by the 3 in this case

Answer 33

I'm debatingif its going to be 0 times sin8x

Answer 34

Well, you can ignore the 3 until the end if you want. Its just in the way as of now, as long as you remember to multiply it in at the end.

Answer 35

theres no way it would be.

Answer 36

oooooh im slow it should've just simply went straight to used the product rule, why im i thinking of the chain rule!

Answer 37

Technically thereis a chain rule everytime you take the derivative of each trig function, but....best not to think of that, just do product rule.

Answer 38

im at cos^2(8x)+3sin^(8x)
without the the 3 in front of cos

Answer 39

I guarantee you should have 8s out there somewhere xD

derivative of sin(8x) = 8cos(8x)
derivative of cos(8x) = -8sin(8x)

Answer 40

oh yeah I wrote down the problem wrong.
So I should've got 8cos^2(8x)-5sin(8x)

Answer 41

How a 5?

Answer 42

I brought down g(X) After addition or was I suppose to ignore the 3 all the way

Answer 43

all the way. Multiply the 3 in at the very end.

Answer 44

ok

Answer 45

so 8cos^2(8x)-8sin^2(8x)

Answer 46

Yep. You can now factor out the 8 and do like what I did above.

Answer 47

im at 24(cos^2(8x)-sin^(8x)
im assuming its gonna be 24cos(16x) like you said

Answer 48

right xDD just because it was the same thing really :P

Answer 49

so for cos-sin is that the trig function and is that how you got cos

Answer 50

its an identity. cos^2x - sin^2x = cos2x.

Answer 51

riiiight. dang i forgot my pre cal stuff :)

Answer 52

thanks a millonsX you just helped me find an easy way of doing stuff I get terrified of of looking at

Answer 53

Yeah, you do it enough and it makes sense thankfully xD OS lagged to death, so sorry for late reply.

Answer 54

no problem