Hi, I have a few derivative problems if anyone is able to help please?

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Hi, I have a few derivative problems if anyone is able to help please?

Calculus1
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|dw:1438832647688:dw|
1) \[f(x)=\tan(x^3)\] I have then used the chain rule as follows (dx of outside) x (inside) x (derivative of inside)
use u substitution

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but we're just taking derivatives, not anti-derivatives where u-sub thrives. o_O
Which I get\[=\sec(x^3) * 3x^2 \]
isn't derivative of tan (x) sec^2x though?
the derivative of x^3 is correct.. it does become 3x^2 and is written outside
So my method is correct but should be \[f \prime (x)=3x^2 \sec^2(x^3)\]?
yeah
awesome, thank you. I have a couple more but not sure where to start as they include \[\ln and \]
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
yes that table needs to be memorized just like all the derivative rules out there x-X!
do you know what the rules are for e? another question is \[f(x)=e^xsinx\]
I thought it would be \[f \prime (x)=e \cos(x)\]
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)
another example |dw:1438833198807:dw|
the derivative of x^2 is written in front of the e but the exponent part where x^2 stays is left alone
Oh ok, sorry just read your response. Great thanks. So in the second question it is only the sin to cos that changes?
yes because we are using product rule on that one \[\LARGE f(x)=e^xsinx\] \[\LARGE f'(x)=e^x(cos(x))+sin(x)e^{x}\]
and the derivative of that x is just one so nothing drastic happens. . .
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
great thank you, could it also be\[e^x(\cos(x))+e^x(\sin(x))\]
yeah... we could factor the e^x to make it nicer, but it's fine the way it is.
order doesn't really matter that much.
cool, makes sense :) Thanks
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?
the derivative of x is one so the derivative of ln(x) is just written as 1/x
maybe quotient rule will work if we had to do \[\LARGE f(x)=\frac{ \ln x }{ x } \]
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
\[\LARGE f(x)=\frac{ \ln x }{ x } \] \[\LARGE f'(x)=\frac{ (x) \frac{1}{x}- ln(x)(1) }{ x^2 } \] \[\LARGE f'(x)=\frac{\frac{x}{x}- ln(x)(1) }{ x^2 } \] \[\LARGE f'(x)=\frac{1- ln(x) }{ x^2 } \]
(fā€™ g āˆ’ gā€™ f )/g2 so I get\[\frac{ 1 }{ x } * x -1 * lnx\] over x^2
yeah we can cancel the x/x and we don't have to write the 1 for the lnx
ok, cool. thanks
So for \[f(x)=e^\sin x\]
as the sin x is an exponent of e it remains untouched?
the latex broke...
\[\LARGE f(x) = e^{sinx}\] like this?
oh, yes
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.
oh, yeah i forgot that we still multiply but do not change the original as well. Cool. Thank you. I have one more question.
\[f(x)=\ln(\sin^2x)\]
would it be\[\frac{ 1 }{ x } *\cos^2\]
wow... ok to make it easier we can re-write that \[\LARGE f(x)=\ln((sinx)^2)\]
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
\[\LARGE f(x)=\ln((sinx)^2) \rightarrow \frac{1}{(sinx)^2}2(sinx)\cos(x)\]
\[\LARGE f'(x) = \frac{2\cos(x)}{\sin(x)}\]
I had one sin x on the top but 2 sin x 's on the bottom.. one of them cancels..
great, that makes much more sense now.
yeah but here's the thing.. .when I checked on wolfram it used a trig identity see that 2sinxcosx ? sin2x = 2sinxcosx
oh
\[\LARGE f'(x)=\ln((sinx)^2) \rightarrow \frac{1}{(sinx)^2}\sin2x\] interesting... we need trig identities. according to mr.wolfram the derivative is 2 cot(x)
hmmm. I got this on mathway 2csc2(x)cos(x)sin(x)
but not sure how it got there....
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
\[\LARGE f(x) = \ln(a) \rightarrow f'(x) = \frac{1}{a}\]
so if we have \[\LARGE a = (sinx)^2 \] \[\LARGE a' = 2(sinx)(cosx) \]
however there is a trig identity so that 2sinxcosx is replaced with sin2x
OMG. ok let's use log rules on this beast.
exponent rule is needed so... \[\LARGE f(x)=\ln((sinx)^2)\] becomes \[\LARGE f(x)=2\ln((sinx))\]
and then use product rule... YEAH I SEE IT NOW!
we just need one part of the product rule because 2 is a constant and all constants have the derivative of 0
\[\LARGE f(x)=2\ln((sinx)) \] \[\LARGE f'(x)=2(\frac{1}{sinx})(cosx)+ \ln(sinx)(0)\]
\[\LARGE f'(x)=2(\frac{cosx}{sinx}) \rightarrow 2\cot(x) \]
sh.it I had that earlier ... just forgot to take the identity of cosx/sinx
12 minutes ago I had this \[\LARGE f'(x) = \frac{2\cos(x)}{\sin(x)} \]
cosx/sinx = cot x
I have not come across trig identities yet, another set of rules for specific trig functions?
oh ok
trigonometry comes before calculus... so eventually by now all the trig identities have to be mastered and memorized... I just overlooked that one part.
\[\LARGE f'(x) = 2 \cot(x) \]
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.
awesome, I will have a closer look at trigonometry. Thank you so much for all of your help. Has been amazing!
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
All good, was so lost without your help. Thanks again, I will close this now :)
ok :)

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