Hi, I have a few derivative problems if anyone is able to help please?
 Carissa15
Hi, I have a few derivative problems if anyone is able to help please?
 Stacey Warren  Expert brainly.com
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 triciaal
dw:1438832647688:dw
 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
use u substitution
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 UsukiDoll
but we're just taking derivatives, not antiderivatives where usub thrives. o_O
 Carissa15
Which I get\[=\sec(x^3) * 3x^2 \]
 UsukiDoll
isn't derivative of tan (x) sec^2x though?
 UsukiDoll
the derivative of x^3 is correct.. it does become 3x^2 and is written outside
 Carissa15
So my method is correct but should be \[f \prime (x)=3x^2 \sec^2(x^3)\]?
 UsukiDoll
yeah
 Carissa15
awesome, thank you. I have a couple more but not sure where to start as they include \[\ln and \]
 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
yes that table needs to be memorized just like all the derivative rules out there xX!
 Carissa15
do you know what the rules are for e? another question is \[f(x)=e^xsinx\]
 Carissa15
I thought it would be \[f \prime (x)=e \cos(x)\]
 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
another example dw:1438833198807:dw
 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
Oh ok, sorry just read your response. Great thanks. So in the second question it is only the sin to cos that changes?
 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
and the derivative of that x is just one so nothing drastic happens. . .
 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
great thank you, could it also be\[e^x(\cos(x))+e^x(\sin(x))\]
 UsukiDoll
yeah... we could factor the e^x to make it nicer, but it's fine the way it is.
 UsukiDoll
order doesn't really matter that much.
 Carissa15
cool, makes sense :) Thanks
 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
the derivative of x is one so the derivative of ln(x) is just written as 1/x
 UsukiDoll
maybe quotient rule will work if we had to do \[\LARGE f(x)=\frac{ \ln x }{ x } \]
 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
\[\LARGE f(x)=\frac{ \ln x }{ x } \]
\[\LARGE f'(x)=\frac{ (x) \frac{1}{x} ln(x)(1) }{ x^2 } \]
\[\LARGE f'(x)=\frac{\frac{x}{x} ln(x)(1) }{ x^2 } \]
\[\LARGE f'(x)=\frac{1 ln(x) }{ x^2 } \]
 Carissa15
(fā g ā gā f )/g2
so I get\[\frac{ 1 }{ x } * x 1 * lnx\] over x^2
 UsukiDoll
yeah we can cancel the x/x
and we don't have to write the 1 for the lnx
 Carissa15
ok, cool. thanks
 Carissa15
So for \[f(x)=e^\sin x\]
 Carissa15
as the sin x is an exponent of e it remains untouched?
 UsukiDoll
the latex broke...
 UsukiDoll
\[\LARGE f(x) = e^{sinx}\]
like this?
 Carissa15
oh, yes
 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
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
\[f(x)=\ln(\sin^2x)\]
 Carissa15
would it be\[\frac{ 1 }{ x } *\cos^2\]
 UsukiDoll
wow... ok to make it easier we can rewrite that
\[\LARGE f(x)=\ln((sinx)^2)\]
 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
\[\LARGE f(x)=\ln((sinx)^2) \rightarrow \frac{1}{(sinx)^2}2(sinx)\cos(x)\]
 UsukiDoll
\[\LARGE f'(x) = \frac{2\cos(x)}{\sin(x)}\]
 UsukiDoll
I had one sin x on the top but 2 sin x 's on the bottom.. one of them cancels..
 Carissa15
great, that makes much more sense now.
 UsukiDoll
yeah but here's the thing.. .when I checked on wolfram it used a trig identity
see that 2sinxcosx ?
sin2x = 2sinxcosx
 Carissa15
oh
 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
hmmm. I got this on mathway 2csc2(x)cos(x)sin(x)
 Carissa15
but not sure how it got there....
 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
\[\LARGE f(x) = \ln(a) \rightarrow f'(x) = \frac{1}{a}\]
 UsukiDoll
so if we have \[\LARGE a = (sinx)^2 \]
\[\LARGE a' = 2(sinx)(cosx) \]
 UsukiDoll
however there is a trig identity
so that 2sinxcosx is replaced with sin2x
 UsukiDoll
OMG. ok let's use log rules on this beast.
 UsukiDoll
exponent rule is needed so...
\[\LARGE f(x)=\ln((sinx)^2)\] becomes
\[\LARGE f(x)=2\ln((sinx))\]
 UsukiDoll
and then use product rule... YEAH I SEE IT NOW!
 UsukiDoll
we just need one part of the product rule because 2 is a constant and all constants have the derivative of 0
 UsukiDoll
\[\LARGE f(x)=2\ln((sinx)) \]
\[\LARGE f'(x)=2(\frac{1}{sinx})(cosx)+ \ln(sinx)(0)\]
 UsukiDoll
\[\LARGE f'(x)=2(\frac{cosx}{sinx}) \rightarrow 2\cot(x) \]
 UsukiDoll
sh.it I had that earlier ... just forgot to take the identity of cosx/sinx
 UsukiDoll
12 minutes ago I had this
\[\LARGE f'(x) = \frac{2\cos(x)}{\sin(x)} \]
 UsukiDoll
cosx/sinx = cot x
 Carissa15
I have not come across trig identities yet, another set of rules for specific trig functions?
 Carissa15
oh ok
 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
\[\LARGE f'(x) = 2 \cot(x) \]
 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
awesome, I will have a closer look at trigonometry. Thank you so much for all of your help. Has been amazing!
 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
All good, was so lost without your help. Thanks again, I will close this now :)
 UsukiDoll
ok :)
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