设z=sinx+f(xy,x^2+y^2),其中x有连续的二阶偏导,求,z对x的偏导和该偏导数对y的二阶偏导
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∂z/∂x
=cosx + f1' * ∂(xy)/∂x + f2' * ∂(x²+y²)/∂x
=cosx + y* f1' +2x *f2'
∂²z/∂x∂y
=∂(cosx + y* f1' +2x *f2') /∂y
= f1' + y* ∂(f1')/∂y + 2x *∂(f2')/∂y
= f1' + y* f11" *∂(xy)/∂y +y* f12" *∂(x²+y²)/∂y +2x* f21" *∂(xy)/∂y+2x* f22" *∂(x²+y²)/∂y
显然
∂(x²+y²)/∂y=2y,而∂(xy)/∂y=x
所以
∂²z/∂x∂y
= f1' + xy* f11" +2y² * f12" + 2x² * f21" + 4xy * f22" (注意f12"= f21" )
= f1' + xy* f11" +(2x² +2y²) * f12" + 4xy * f22"
=cosx + f1' * ∂(xy)/∂x + f2' * ∂(x²+y²)/∂x
=cosx + y* f1' +2x *f2'
∂²z/∂x∂y
=∂(cosx + y* f1' +2x *f2') /∂y
= f1' + y* ∂(f1')/∂y + 2x *∂(f2')/∂y
= f1' + y* f11" *∂(xy)/∂y +y* f12" *∂(x²+y²)/∂y +2x* f21" *∂(xy)/∂y+2x* f22" *∂(x²+y²)/∂y
显然
∂(x²+y²)/∂y=2y,而∂(xy)/∂y=x
所以
∂²z/∂x∂y
= f1' + xy* f11" +2y² * f12" + 2x² * f21" + 4xy * f22" (注意f12"= f21" )
= f1' + xy* f11" +(2x² +2y²) * f12" + 4xy * f22"
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