Question

Use the appropriate rule or combination of rules to find the derivative of the function defined below.
y = sin (x^4) cos (x^4)

Answer 1

i used the product rule for both sin and cos to get
(sin*4x^3+x^4*cos)(cos*4x^3*x^4-sinx)

Answer 2

but my answer should look something like: 4cos(x^4)^2*x^4 -4sin(x^4)^2*x^4

Answer 3

product rule and chain rule

Answer 4

first differentiate sin(x^4) and then x^4.. do the same for cos also

Answer 5

the answer you wrote second is not right either
you need \((fg)'=f'g+g'f\) with
\[f(x)=\sin(x^4), f'(x)=\cos(x^4)\times 4x^3=4x^3\cos(x^4)\]

Answer 6

similarly if \(g(x)=\cos(x^4)\) then \(g'(x)=-\sin(x^4)\times 4x^3=-4x^3\sin(x^4)\)

Answer 7

@satellite73 said... there will be no x^4 term it is x^3 term

Answer 8

i thought that sin and x^4 were two different factors.

Answer 9

?

Answer 10

should i end up with something that looks like: \[4x^3*\cos(x^4)*\cos(x^4)+(-4sinx^3*\sin(x^4))*\sin(x^4)\]

Answer 11

ok for simplicity.. do one thing..
let some y = x^4

Answer 12

now your equation will be sin(y)cos(y)..

Answer 13

ok

Answer 14

so i get: siny'cosy+cosy'-siny

Answer 15

you know how to differentiate this right.. that will (cos(y)^2-sin(y)^2)dy

Answer 16

yes

Answer 17

i do the product as: fg'+gf'

Answer 18

yes perfect.. d(uv) =vd(u)+ud(v) right.. v =cos(y) u=sin(y).. from this you will get the above thing

Answer 19

i don't see how you're getting it to the power of 2 when it should be 1-1. shouldnt it?

Answer 20

now our y = x^4.. so y'= (4x^3) x'

Answer 21

put y and y' in above equation.. what do you get.. 4x^3(cos(x^4)^2-sin(x^4)^2)

Answer 22

i didn't get everything to the power of 2, i just had(sin(4x^3)+cos(x^4))+((cos(4x^3)-sin(x^4)

Answer 23

what you have seems to look like that answer though is it?

Answer 24

ok.. how did you get 4terms?

Answer 25

mistake you are would probably be this.. you are combining both product rule and chain rule..

Answer 26

i think that is what i'm doing,
i thought that is what i neeeded to do.

so my answer should be: 4x^3(cos(x^4)^2-sin(x^4)^2)

Answer 27

see when you differentiate sin(x^4).. first you sin derivative (cos(x^4)*(derivative of x^4))this whole thing is multiplied by cos(x^4).. do the same for cos(x^4) also

Answer 28

yes i have that done.

Answer 29

is 4x^3(cos(x^4)^2-sin(x^4)^2) what i should have?
i find it easier to have a correct answer so i can look through my workings and see what i've done wrong.

Answer 30

yes.. that is what you should get..

Answer 31

can i write it as: 4x^3*cos(x^4)^2-sin(x^4)^2