At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

|dw:1438832647688:dw|

use u substitution

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 \[\ln and \]

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 \[f \prime (x)=e \cos(x)\]

another example |dw:1438833198807:dw|

and the derivative of that x is just one so nothing drastic happens. . .

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

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 \[\LARGE f(x)=\frac{ \ln x }{ x } \]

(fâ€™ g âˆ’ gâ€™ 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

\[f(x)=\ln(\sin^2x)\]

would it be\[\frac{ 1 }{ x } *\cos^2\]

wow... ok to make it easier we can re-write that
\[\LARGE f(x)=\ln((sinx)^2)\]

\[\LARGE f(x)=\ln((sinx)^2) \rightarrow \frac{1}{(sinx)^2}2(sinx)\cos(x)\]

\[\LARGE f'(x) = \frac{2\cos(x)}{\sin(x)}\]

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.

oh

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!

\[\LARGE f(x)=2\ln((sinx)) \]
\[\LARGE f'(x)=2(\frac{1}{sinx})(cosx)+ \ln(sinx)(0)\]

\[\LARGE f'(x)=2(\frac{cosx}{sinx}) \rightarrow 2\cot(x) \]

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

\[\LARGE f'(x) = 2 \cot(x) \]

All good, was so lost without your help. Thanks again, I will close this now :)

ok :)