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

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

product rule and chain rule

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

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

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

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

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

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

ok

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

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

yes

i do the product as: fg'+gf'

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

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

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

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

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)

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

ok.. how did you get 4terms?

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

yes i have that done.

yes.. that is what you should get..

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