Question

find the derivative of the following...Using the chain rule
f(x)=square root sin(ln(4x^3+1))

Answer 1

\[f(x)=\sqrt{\sin(\ln(4x^3+1))}\]

Answer 2

the derivative of \(\sqrt{f(x)}\) is
\[\frac{1}{2\sqrt{f(x)}}\] so the first step is
\[\frac{1}{2\sqrt{\sin(\ln(4x^2+1))}}\times \frac{d}{dx}[\sin(\ln(4x^2+1))]\]\]

Answer 3

the first line i wrote was wrong
it should be
the derivative of \(\sqrt{f(x)}\) is
\[\frac{f'(x)}{2\sqrt{f(x)}}\]

Answer 4

then the derivative of sine is cosine, so next step is \[\frac{1}{2\sqrt{\sin(\ln(4x^2+1))}}\times \cos(\ln(4x^2+1))\times\frac{d}{dx}[\ln(4x^2+1)\]

Answer 5

ok got it

Answer 6

ok good
two more steps

Answer 7

ok I am stuck I am ghalf way through and now stuck

Answer 8

ok we got the first part right?

Answer 9

yes I understood that

Answer 10

last job is
\[\frac{d}{dx}[\ln(4x^2+1)]\]

Answer 11

\[\frac{d}{dx}\ln(f(x))=\frac{f'(x)}{f(x)}\]

Answer 12

so last step is
\[\frac{8x}{4x^2+1}\]

Answer 13

chain rule all the way down

Answer 14

Ok then I got \[\frac{ -32x^2+8 }{ (4x^2+1)^2}\]

Answer 15

hold the phone
where did the numerator come from?

Answer 16

you need the derivative of \(\ln(4x^2+1)\)
that is all

Answer 17

the derivative of \(\ln(4x^2+1)\) is \(\frac{8x}{4x^2+1}\)
you do not need to take the derivative of that, it IS the derivative

Answer 18

\[\frac{1}{2\sqrt{\sin(\ln(4x^2+1))}}\times \cos(\ln(4x^2+1))\times\frac{d}{dx}[\ln(4x^2+1)\] last step is
\[\frac{d}{dx}[\ln(4x^2+1)]=\frac{8x}{4x^2+1}\] and you are done

Answer 19

ok so then the (8x)/4x^2+1) is the answer

Answer 20

to that part, yes'

Answer 21

entire answer is
\[\frac{1}{2\sqrt{\sin(\ln(4x^2+1))}}\times \cos(\ln(4x^2+1))\times\frac{8x}{4x^2+1}\]

Answer 22

oh gotcha why does the I was putting in the wrong order on my paper writing it out is why it wasn't making any sense sorry:)

Answer 23

not too bad really, peal off the functions one at a time