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

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

Calculus1
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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

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Other answers:

first differentiate sin(x^4) and then x^4.. do the same for cos also
the answer you wrote second is not right either you need \((fg)'=f'g+g'f\) with \[f(x)=\sin(x^4), f'(x)=\cos(x^4)\times 4x^3=4x^3\cos(x^4)\]
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.
?
should i end up with something that looks like: \[4x^3*\cos(x^4)*\cos(x^4)+(-4sinx^3*\sin(x^4))*\sin(x^4)\]
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..
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)
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
yes i have that done.
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.
yes.. that is what you should get..
can i write it as: 4x^3*cos(x^4)^2-sin(x^4)^2

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