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/sci/ - Science & Math

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5471922 No.5471922 [Reply] [Original]

Uploaded for a friend, but if you guys find any errors that'd be nice since I just retaught myself log differentiation

>> No.5471943

Write dy/dx you lazy fuck.

>> No.5471948
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5471948

>>5471922
>derivative of log(e^(sinx)) = sin(x)

I see this mistake all the time with calc 1 freshman and I don't know why this pisses me off so much.

IT'S A GODDAMN IDENTITY NOT THE DERIVATIVE DAMNIT

Same thing with the trig functions, but in the opposite direction.
>"So then what does tan(x) mean?"
>"Oh that's sec^2(x)."

>> No.5471986

>>5471948
I've never seen anyone make that mistake ever. That's really bad. Hahahaha, well at least I guess it means OP isn't the only one.

OP, read up on your logarithms. Possibly start working your problems with an extra step where you distribute the differential to every item you're differentiating so that it's easier to spot your mistakes (on your own).

>> No.5472028

so (dy/dx) ln (e^(sinx)) should be what then? this is literally the first log diff. equation i've worked on in years so excuse my ignorance

My guess is ln (e^(sinx) * (sinx))

>> No.5472031

>>5472028
ln(e^(sinx)) = sinx

d/dx (ln(e^sinx)) = d/dx (sinx)

Finish the rest.

>> No.5472032

>>5472028
>log(e^(sinx))

d/dx log(e^(sin(x)) = d/dx sin(x)
= cos(x)

>> No.5472033

>>5472028
>ln (e^(sinx) * (sinx))

Meant to say ln (e^(sinx) * (cosx) )

>> No.5472036

Where the fuck did that 3 come from on the last term on the second step?

>> No.5472038

>>5472033

Still incorrect.

>> No.5472042

>>5472036
yeah i just corrected that as i wrote it down, in the original problem its supposed to be e^sinx * (5x^2 + x - 1)^3 in the denom. i transfered it over wrong when i switched sides to write it cleaner

>> No.5472043

>>5471922
Also you forgot chain rule on the derivative of cos^2(3x), the three. And why did you subtract the sin(3x) are you retarded?

>> No.5472049

OP, thing is that the natural logarithm and e are inverses. If you have e^x (doesn't matter what x is), and you take the natural logarithm of it such that ln(e^x) then it resolves to just x. You do this BEFORE you differentiate. I'm sure you can differentiate x without any problem.

>> No.5472063

>>5472043
cos^2 x = (cosx)(cosx) = 2cosx - sinx

so i guess it should be

cos^(2) 3x = (cosx)(cosx)(3) = (2cosx - sinx)(3) = 6cosx - 3sinx

But wouldn't that be the same thing as 2cos(3x) - sin(3x) still? I know its wrong to write it like that now that it's been pointed out

>> No.5472100

>>5472063
cos^2(3x) = -6cos(3x)sin(3x)
Use power rule for the two, chain rule for the cos(3x) which tell us to multiply by sin(3x), but then there's a 3 in the parenthesis, so we have to do chain rule again, which tells us to multiply by the derivative of (3x) which is 3. So, you multiply the original 2 by 3 to get 6, multiply it by the sin(3x), and now you're done.

>> No.5472101

>>5472100
>>-sin(3x)
typo

>> No.5472104

>>5472063
Listen, just by what you've typed you aren't ready for calculus or w.e. the problem is. That three can't come outside of the cosine, it's a multiplying factor for the argument inside. You need to learn the prerequisites

sin(3x) =/= 3sin(x)