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Question-133980

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Question Number 133980 by mohammad17 last updated on 26/Feb/21
Answered by mr W last updated on 26/Feb/21
(∂p/∂x)=(x/( (√(x^2 +y^2 +z^2 )))) (∂p/∂y)=(y/( (√(x^2 +y^2 +z^2 )))) (∂p/∂z)=(z/( (√(x^2 +y^2 +z^2 )))) (∂w/∂x)=(dw/dp)×(∂p/∂x)=(dw/dp)×(x/( (√(x^2 +y^2 +z^2 )))) ((∂w/∂x))^2 =((dw/dp))^2 ×(x^2 /( x^2 +y^2 +z^2 )) ((∂w/∂y))^2 =((dw/dp))^2 ×(y^2 /( x^2 +y^2 +z^2 )) ((∂w/∂z))^2 =((dw/dp))^2 ×(z^2 /( x^2 +y^2 +z^2 )) ⇒((∂w/∂x))^2 +((∂w/∂y))^2 +((∂w/∂z))^2 =((dw/dp))^2
∂p∂x=xx2+y2+z2∂p∂y=yx2+y2+z2∂p∂z=zx2+y2+z2∂w∂x=dwdp×∂p∂x=dwdp×xx2+y2+z2Missing \left or extra \rightMissing \left or extra \rightMissing \left or extra \rightMissing \left or extra \right

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