Question

find the derivative of the function.

Answer 1

\[y=\sin (\sqrt[3]{x})+\sqrt[3]{\sin(4x)}\]

Answer 2

chain rule is definitely needed

Answer 3

\[y=\sin(x^\frac{1}{3})+(\sin(4x))^\frac{1}{3}\]

Answer 4

yes i have that...i just don't know where to go from there

Answer 5

lets look at the first function
the inside is x^(1/3)
how do you differentiate that?.

Answer 6

1/3x

Answer 7

don't forget to subtract one from the exponent

Answer 8

1/3x^2/3

Answer 9

that is \[\frac{1}{3x^\frac{2}{3}}\]

correct?

Answer 10

yes

Answer 11

ok let's go to the outside function that was there
what is derivative of sin?

Answer 12

cos?

Answer 13

\[[\sin(x^\frac{1}{3})]'=(x^\frac{1}{3})'\cos(x^\frac{1}{3})=\frac{1}{3x^\frac{2}{3} }\cos(x^\frac{1}{3})\]

Answer 14

so let's go to the other function

Answer 15

the very inside is 4x
what is derivative of 4x?

Answer 16

4

Answer 17

then go to the middle inside
again the derivative of sin is cos

Answer 18

then finanlly you need to differentiate that ( )^(1/3) part

Answer 19

would it be 4cos^2/3?

Answer 20

leave the inside what it is

Answer 21

differentiate x^(1/3)

Answer 22

the inside will stay the same

Answer 23

so i'll end up with:
\[\frac{ 1 }{ 3x ^{2/3} }\cos(x ^{1/3})+4\cos(x ^{1/3})\]

Answer 24

\[[(\sin(4x))^\frac{1}{3}]'=(4x)' \cos(4x) \frac{1}{3}(\sin(4x))^{\frac{1}{3}-1} \\ =4 \cos(4x)\frac{1}{3}(\sin(4x))^{\frac{1}{3}-1} \]
do you see did derivative of very inside 4x
then we went to the middle inside sin
then we go to the very outside ( )^(1/3))

Answer 25

oh ok..i see

Answer 26

so say we have
to differentiate
\[\sin(\cos(\sec(4x)))\] with respect to x

we get
\[4 \cdot \sec(4x) \tan(4x) \cdot -\sin(\sec(4x)) \cdot \cos(\cos(\sec(4x))\]
as we work from inside to outside
make sure you keep the same insides as you see here

Answer 27

derivative of 4x is 4
derivative of sec( ) is sec()tan()
derivative of cos() is -sin( )
derivative of sin( ) is cos()

Answer 28

do you see how i moved from inside to outside
I kept the inside functions for each of the inside to outside functions

Answer 29

yes

Answer 30

say we wanted to differentiate \[\sqrt{1-(\sin(\cos(x)))^2}\]
the inside is 1-(sin(cos(x)))^2
so we start differentiating that first
derivative of 1 is 0 of course
but we need chain rule for this other part of the inside function
so derivative of cos(x) is -sin(x)
derivative of sin( ) is cos()
derivative of ()^2 is 2()

then we have the outside function was ()^(1/2)
derivative of that is (1/2)( )^(-1/2)
so we have
\[[0--\sin(x)\cdot \cos(\cos(x)) \cdot 2(\sin(\cos(x))] \cdot \frac{1}{2}(1-(\sin(\cos(x)))^2)^\frac{-1}{2}\]

Answer 31

ok, i see...