anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
i used the product rule for both sin and cos to get (sin*4x^3+x^4*cos)(cos*4x^3*x^4-sinx)
anonymous
  • anonymous
but my answer should look something like: 4cos(x^4)^2*x^4 -4sin(x^4)^2*x^4
anonymous
  • anonymous
product rule and chain rule

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anonymous
  • anonymous
first differentiate sin(x^4) and then x^4.. do the same for cos also
anonymous
  • anonymous
the answer you wrote second is not right either you need \((fg)'=f'g+g'f\) with \[f(x)=\sin(x^4), f'(x)=\cos(x^4)\times 4x^3=4x^3\cos(x^4)\]
anonymous
  • anonymous
similarly if \(g(x)=\cos(x^4)\) then \(g'(x)=-\sin(x^4)\times 4x^3=-4x^3\sin(x^4)\)
anonymous
  • anonymous
@satellite73 said... there will be no x^4 term it is x^3 term
anonymous
  • anonymous
i thought that sin and x^4 were two different factors.
anonymous
  • anonymous
?
anonymous
  • anonymous
should i end up with something that looks like: \[4x^3*\cos(x^4)*\cos(x^4)+(-4sinx^3*\sin(x^4))*\sin(x^4)\]
anonymous
  • anonymous
ok for simplicity.. do one thing.. let some y = x^4
anonymous
  • anonymous
now your equation will be sin(y)cos(y)..
anonymous
  • anonymous
ok
anonymous
  • anonymous
so i get: siny'cosy+cosy'-siny
anonymous
  • anonymous
you know how to differentiate this right.. that will (cos(y)^2-sin(y)^2)dy
anonymous
  • anonymous
yes
anonymous
  • anonymous
i do the product as: fg'+gf'
anonymous
  • anonymous
yes perfect.. d(uv) =vd(u)+ud(v) right.. v =cos(y) u=sin(y).. from this you will get the above thing
anonymous
  • anonymous
i don't see how you're getting it to the power of 2 when it should be 1-1. shouldnt it?
anonymous
  • anonymous
now our y = x^4.. so y'= (4x^3) x'
anonymous
  • anonymous
put y and y' in above equation.. what do you get.. 4x^3(cos(x^4)^2-sin(x^4)^2)
anonymous
  • anonymous
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)
anonymous
  • anonymous
what you have seems to look like that answer though is it?
anonymous
  • anonymous
ok.. how did you get 4terms?
anonymous
  • anonymous
mistake you are would probably be this.. you are combining both product rule and chain rule..
anonymous
  • anonymous
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)
anonymous
  • anonymous
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
anonymous
  • anonymous
yes i have that done.
anonymous
  • anonymous
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.
anonymous
  • anonymous
yes.. that is what you should get..
anonymous
  • anonymous
can i write it as: 4x^3*cos(x^4)^2-sin(x^4)^2

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