Carissa15
  • Carissa15
Hi, I have a few derivative problems if anyone is able to help please?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
triciaal
  • triciaal
|dw:1438832647688:dw|
Carissa15
  • Carissa15
1) \[f(x)=\tan(x^3)\] I have then used the chain rule as follows (dx of outside) x (inside) x (derivative of inside)
triciaal
  • triciaal
use u substitution

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UsukiDoll
  • UsukiDoll
but we're just taking derivatives, not anti-derivatives where u-sub thrives. o_O
Carissa15
  • Carissa15
Which I get\[=\sec(x^3) * 3x^2 \]
UsukiDoll
  • UsukiDoll
isn't derivative of tan (x) sec^2x though?
UsukiDoll
  • UsukiDoll
the derivative of x^3 is correct.. it does become 3x^2 and is written outside
Carissa15
  • Carissa15
So my method is correct but should be \[f \prime (x)=3x^2 \sec^2(x^3)\]?
UsukiDoll
  • UsukiDoll
yeah
Carissa15
  • Carissa15
awesome, thank you. I have a couple more but not sure where to start as they include \[\ln and \]
UsukiDoll
  • UsukiDoll
trig derivatives sin (x) -> cos(x) cos (x) -> -sin(x) tan (x) -> sec^2(x) csc(x) - > -csc(x)cot(x) sec(x) -> secxtanx cot(x) = -csc^2x
UsukiDoll
  • UsukiDoll
yes that table needs to be memorized just like all the derivative rules out there x-X!
Carissa15
  • Carissa15
do you know what the rules are for e? another question is \[f(x)=e^xsinx\]
Carissa15
  • Carissa15
I thought it would be \[f \prime (x)=e \cos(x)\]
UsukiDoll
  • UsukiDoll
yeah... usually for e^x the exponent part for the e is left alone so for example the derivative of e^x is just e^x due to the fact that the derivative of x is 1, but nobody writes the 1 so derivative of x is one , but the exponent part that's attached to the e stays e^{x}(1)
UsukiDoll
  • UsukiDoll
another example |dw:1438833198807:dw|
UsukiDoll
  • UsukiDoll
the derivative of x^2 is written in front of the e but the exponent part where x^2 stays is left alone
Carissa15
  • Carissa15
Oh ok, sorry just read your response. Great thanks. So in the second question it is only the sin to cos that changes?
UsukiDoll
  • UsukiDoll
yes because we are using product rule on that one \[\LARGE f(x)=e^xsinx\] \[\LARGE f'(x)=e^x(cos(x))+sin(x)e^{x}\]
UsukiDoll
  • UsukiDoll
and the derivative of that x is just one so nothing drastic happens. . .
UsukiDoll
  • UsukiDoll
I left the e^x alone first and took the derivative of sinx + I left sin x alone and took the derivative of e^x derivative of e^x in that case is just e^x(1) or just written as e^x
Carissa15
  • Carissa15
great thank you, could it also be\[e^x(\cos(x))+e^x(\sin(x))\]
UsukiDoll
  • UsukiDoll
yeah... we could factor the e^x to make it nicer, but it's fine the way it is.
UsukiDoll
  • UsukiDoll
order doesn't really matter that much.
Carissa15
  • Carissa15
cool, makes sense :) Thanks
Carissa15
  • Carissa15
I also have \[f(x)=\frac{ \ln x }{ x }\] I know that the dx of \[\ln (x)\] is \[\frac{ 1 }{ x }\] but not sure what to then do about the x underneath?
UsukiDoll
  • UsukiDoll
the derivative of x is one so the derivative of ln(x) is just written as 1/x
UsukiDoll
  • UsukiDoll
maybe quotient rule will work if we had to do \[\LARGE f(x)=\frac{ \ln x }{ x } \]
UsukiDoll
  • UsukiDoll
so the bottom gets squared then leave the second part (The denominator x alone) deal with ln x the derivative of ln x is 1/x(1) but again since the derivative of x is just 1 we don't have to write that extra stuff.. just 1/x will be good enough. - leave the first part (the numerator ln x alone ) deal with the x the derivative of x is just 1
UsukiDoll
  • UsukiDoll
\[\LARGE f(x)=\frac{ \ln x }{ x } \] \[\LARGE f'(x)=\frac{ (x) \frac{1}{x}- ln(x)(1) }{ x^2 } \] \[\LARGE f'(x)=\frac{\frac{x}{x}- ln(x)(1) }{ x^2 } \] \[\LARGE f'(x)=\frac{1- ln(x) }{ x^2 } \]
Carissa15
  • Carissa15
(fā€™ g āˆ’ gā€™ f )/g2 so I get\[\frac{ 1 }{ x } * x -1 * lnx\] over x^2
UsukiDoll
  • UsukiDoll
yeah we can cancel the x/x and we don't have to write the 1 for the lnx
Carissa15
  • Carissa15
ok, cool. thanks
Carissa15
  • Carissa15
So for \[f(x)=e^\sin x\]
Carissa15
  • Carissa15
as the sin x is an exponent of e it remains untouched?
UsukiDoll
  • UsukiDoll
the latex broke...
UsukiDoll
  • UsukiDoll
\[\LARGE f(x) = e^{sinx}\] like this?
Carissa15
  • Carissa15
oh, yes
UsukiDoll
  • UsukiDoll
so take the derivative of sin(x) and put it in front of the e. leave the exponent part of e which is at e^{sinx} alone.
Carissa15
  • Carissa15
oh, yeah i forgot that we still multiply but do not change the original as well. Cool. Thank you. I have one more question.
Carissa15
  • Carissa15
\[f(x)=\ln(\sin^2x)\]
Carissa15
  • Carissa15
would it be\[\frac{ 1 }{ x } *\cos^2\]
UsukiDoll
  • UsukiDoll
wow... ok to make it easier we can re-write that \[\LARGE f(x)=\ln((sinx)^2)\]
UsukiDoll
  • UsukiDoll
then we know that since we have a log it will be 1/(sinx)^2 but then we have to take the derivative of (sinx)^2
UsukiDoll
  • UsukiDoll
\[\LARGE f(x)=\ln((sinx)^2) \rightarrow \frac{1}{(sinx)^2}2(sinx)\cos(x)\]
UsukiDoll
  • UsukiDoll
\[\LARGE f'(x) = \frac{2\cos(x)}{\sin(x)}\]
UsukiDoll
  • UsukiDoll
I had one sin x on the top but 2 sin x 's on the bottom.. one of them cancels..
Carissa15
  • Carissa15
great, that makes much more sense now.
UsukiDoll
  • UsukiDoll
yeah but here's the thing.. .when I checked on wolfram it used a trig identity see that 2sinxcosx ? sin2x = 2sinxcosx
Carissa15
  • Carissa15
oh
UsukiDoll
  • UsukiDoll
\[\LARGE f'(x)=\ln((sinx)^2) \rightarrow \frac{1}{(sinx)^2}\sin2x\] interesting... we need trig identities. according to mr.wolfram the derivative is 2 cot(x)
Carissa15
  • Carissa15
hmmm. I got this on mathway 2csc2(x)cos(x)sin(x)
Carissa15
  • Carissa15
but not sure how it got there....
UsukiDoll
  • UsukiDoll
ok hold up let's take the (sinx)^2 out for a bit let let a = (sinx)^2 so we just have ln a for now
UsukiDoll
  • UsukiDoll
\[\LARGE f(x) = \ln(a) \rightarrow f'(x) = \frac{1}{a}\]
UsukiDoll
  • UsukiDoll
so if we have \[\LARGE a = (sinx)^2 \] \[\LARGE a' = 2(sinx)(cosx) \]
UsukiDoll
  • UsukiDoll
however there is a trig identity so that 2sinxcosx is replaced with sin2x
UsukiDoll
  • UsukiDoll
OMG. ok let's use log rules on this beast.
UsukiDoll
  • UsukiDoll
exponent rule is needed so... \[\LARGE f(x)=\ln((sinx)^2)\] becomes \[\LARGE f(x)=2\ln((sinx))\]
UsukiDoll
  • UsukiDoll
and then use product rule... YEAH I SEE IT NOW!
UsukiDoll
  • UsukiDoll
we just need one part of the product rule because 2 is a constant and all constants have the derivative of 0
UsukiDoll
  • UsukiDoll
\[\LARGE f(x)=2\ln((sinx)) \] \[\LARGE f'(x)=2(\frac{1}{sinx})(cosx)+ \ln(sinx)(0)\]
UsukiDoll
  • UsukiDoll
\[\LARGE f'(x)=2(\frac{cosx}{sinx}) \rightarrow 2\cot(x) \]
UsukiDoll
  • UsukiDoll
sh.it I had that earlier ... just forgot to take the identity of cosx/sinx
UsukiDoll
  • UsukiDoll
12 minutes ago I had this \[\LARGE f'(x) = \frac{2\cos(x)}{\sin(x)} \]
UsukiDoll
  • UsukiDoll
cosx/sinx = cot x
Carissa15
  • Carissa15
I have not come across trig identities yet, another set of rules for specific trig functions?
Carissa15
  • Carissa15
oh ok
UsukiDoll
  • UsukiDoll
trigonometry comes before calculus... so eventually by now all the trig identities have to be mastered and memorized... I just overlooked that one part.
UsukiDoll
  • UsukiDoll
\[\LARGE f'(x) = 2 \cot(x) \]
UsukiDoll
  • UsukiDoll
so I didn't have to use sin2x = 2sinx cosx after all... just had to cancel one of the sinx and then use cosx/sinx = cotx and that's what wolfram got too.
Carissa15
  • Carissa15
awesome, I will have a closer look at trigonometry. Thank you so much for all of your help. Has been amazing!
UsukiDoll
  • UsukiDoll
yeah... it wasn't too bad... except for the last one when I overlooked something... XD! and then taking the identity of 2sinxcosx made it worse. XD
Carissa15
  • Carissa15
All good, was so lost without your help. Thanks again, I will close this now :)
UsukiDoll
  • UsukiDoll
ok :)

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