求y = (sinx/4)^4 + (cosx/4)^4导数,
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求y = (sinx/4)^4 + (cosx/4)^4导数,
是sin(x/4)?
y = (sin(x/4))^4 + (cos(x/4))^4
y'=4*(sin(x/4))^3*(sin(x/4))'+4*(cos(x/4))^3*(cos(x/4))'
=4*(sin(x/4))^3*cos(x/4)*1/4+4*(cos(x/4))^3*(-sin(x/4))*1/4
=(sin(x/4))^3*cos(x/4)-(cos(x/4))^3*sin(x/4)
=sin(x/4)*cos(x/4)*((sin(x/4))^2-(cos(x/4))^2)
=1/2*sin(x/2)*(-cos(x/2))
=-1/4*sinx
或者
y=(sin(x/4))^4 + (cos(x/4))^4
=(sin(x/4))^4 + (cos(x/4))^4+2(sin(x/4)*cos(x/4))^2-2(sin(x/4)*cos(x/4))^2
=((sin(x/4))^2+(cos(x/4))^2)^2-1/2*(sin(x/2))^2
=1-1/4*(1-cosx)
=1/4*cosx
所以
y'=-1/4*sinx
y = (sin(x/4))^4 + (cos(x/4))^4
y'=4*(sin(x/4))^3*(sin(x/4))'+4*(cos(x/4))^3*(cos(x/4))'
=4*(sin(x/4))^3*cos(x/4)*1/4+4*(cos(x/4))^3*(-sin(x/4))*1/4
=(sin(x/4))^3*cos(x/4)-(cos(x/4))^3*sin(x/4)
=sin(x/4)*cos(x/4)*((sin(x/4))^2-(cos(x/4))^2)
=1/2*sin(x/2)*(-cos(x/2))
=-1/4*sinx
或者
y=(sin(x/4))^4 + (cos(x/4))^4
=(sin(x/4))^4 + (cos(x/4))^4+2(sin(x/4)*cos(x/4))^2-2(sin(x/4)*cos(x/4))^2
=((sin(x/4))^2+(cos(x/4))^2)^2-1/2*(sin(x/2))^2
=1-1/4*(1-cosx)
=1/4*cosx
所以
y'=-1/4*sinx
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