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

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bmelykBest ResponseYou've already chosen the best response.0
i used the product rule for both sin and cos to get (sin*4x^3+x^4*cos)(cos*4x^3*x^4sinx)
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
but my answer should look something like: 4cos(x^4)^2*x^4 4sin(x^4)^2*x^4
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
product rule and chain rule
 one year ago

theloserBest ResponseYou've already chosen the best response.0
first differentiate sin(x^4) and then x^4.. do the same for cos also
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
the answer you wrote second is not right either you need \((fg)'=f'g+g'f\) with \[f(x)=\sin(x^4), f'(x)=\cos(x^4)\times 4x^3=4x^3\cos(x^4)\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
similarly if \(g(x)=\cos(x^4)\) then \(g'(x)=\sin(x^4)\times 4x^3=4x^3\sin(x^4)\)
 one year ago

theloserBest ResponseYou've already chosen the best response.0
@satellite73 said... there will be no x^4 term it is x^3 term
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
i thought that sin and x^4 were two different factors.
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
should i end up with something that looks like: \[4x^3*\cos(x^4)*\cos(x^4)+(4sinx^3*\sin(x^4))*\sin(x^4)\]
 one year ago

theloserBest ResponseYou've already chosen the best response.0
ok for simplicity.. do one thing.. let some y = x^4
 one year ago

theloserBest ResponseYou've already chosen the best response.0
now your equation will be sin(y)cos(y)..
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
so i get: siny'cosy+cosy'siny
 one year ago

theloserBest ResponseYou've already chosen the best response.0
you know how to differentiate this right.. that will (cos(y)^2sin(y)^2)dy
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
i do the product as: fg'+gf'
 one year ago

theloserBest ResponseYou've already chosen the best response.0
yes perfect.. d(uv) =vd(u)+ud(v) right.. v =cos(y) u=sin(y).. from this you will get the above thing
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
i don't see how you're getting it to the power of 2 when it should be 11. shouldnt it?
 one year ago

theloserBest ResponseYou've already chosen the best response.0
now our y = x^4.. so y'= (4x^3) x'
 one year ago

theloserBest ResponseYou've already chosen the best response.0
put y and y' in above equation.. what do you get.. 4x^3(cos(x^4)^2sin(x^4)^2)
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
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)
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
what you have seems to look like that answer though is it?
 one year ago

theloserBest ResponseYou've already chosen the best response.0
ok.. how did you get 4terms?
 one year ago

theloserBest ResponseYou've already chosen the best response.0
mistake you are would probably be this.. you are combining both product rule and chain rule..
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
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)^2sin(x^4)^2)
 one year ago

theloserBest ResponseYou've already chosen the best response.0
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
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
is 4x^3(cos(x^4)^2sin(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.
 one year ago

theloserBest ResponseYou've already chosen the best response.0
yes.. that is what you should get..
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
can i write it as: 4x^3*cos(x^4)^2sin(x^4)^2
 one year ago
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