Solution:
1. This expression can be written as:
d2xd2y(csc(x))=dxdy(dxdycsc(x))
2. Find the first derivative of the specified function:
dxdycsc(x)=dxdy(sin(x)1)=sin2(x)(1′)×sin(x)−1×(sin(x))′=−sin2(x)cos(x)
3. Next, find the derivative of the resulting function:
dxdy(sin2(x)−cos(x))=(sin2(x))2(−cos(x))′×sin2(x)−(−cos(x))×(sin2(x))′=
=sin4(x)sin(x)×sin2(x)+cos(x)×(2sin(x)cos(x))=sin4(x)sin(x)×(sin2(x)+2cos2(x))=
=sin3(x)sin2(x)+cos2(x)+cos2(x)=sin3(x)1+cos2(x)=sin3(x)1(1+cos2(x))=
=csc3(x)(1+cos2(x))
Answer:
d2xd2y(csc(x))=(cos2(x)+1)csc3(x)
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