Question

Find derivative of (2x+1)^3 sin(4x)

Answer 1

product and chain for this one

Answer 2

\[\left(fg\right)'=f'g+g'f\]with
\[f(x)=)2x+1)^3,f'(x)=3(2x+1)^2\times 2, g(x)=\sin(4x), g'(x)=4\cos(4x)\]

Answer 3

some kind of typo there, but maybe it is clear

Answer 4

\[f(x)=(2x+1)^3,f'(x)=3(2x+1)^2\times 2\]\[ g(x)=\sin(4x), g'(x)=4\cos(4x)\]

Answer 5

you good from there?

Answer 6

Why is there a x2 at the end?

Answer 7

because the derivative of \(2x+1\) is \(2\)

Answer 8

really it is
\[f'(x)=6(2x+1)^2\]

Answer 9

Ohh OK! Do you think you can help me with one more?

Answer 10

sure why not

Answer 11

you still have to put all this together using the product rule, but that is just a matter of writing it

Answer 12

you can post the next one here if you like

Answer 13

OK. The other one is (2x+1)^3 tan(x)

Answer 14

really? it is almost identical to this one
again you need the product rule, which i hope you know \[\left(fg\right)'=f'g+g'f\]

Answer 15

But doesn't the tan make it different?

Answer 16

this time
\[f(x)=(2x+1)^3,f'(x)=6(2x+1)^2\]
\[g(x)=\tan(x), g'(x)=\sec^2(x)\] yeah it makes a difference, the derivative of tangent is secant squared, not cosine!

Answer 17

but the formula and the work is still the same

Answer 18

Got it from there. Thanks for your patience! Haha I'M kinda slow

Answer 19

you do understand you have to put all that together in this
\[\left(fg\right)'=f'g+g'f\] right?

Answer 20

Less. My Cow assignment will only take it in that form

Answer 21

slow schmo
you were not born knowing this
it all takes some practice
then you can forget all about it later

Answer 22

*yes

Answer 23

COW as in temple's cow?

Answer 24

Yes

Answer 25

you go there? or just use the system?

Answer 26

Our teacher has us just use the system

Answer 27

oh wow it is ancient
surprised they don't use mylabs plus or webassign
hope dan reich is making some money out of this

Answer 28

Well I do live in south Dakota aka the middle of nowhere so, figures. Haha