1. Given z=ax+by then we will have the following: ∂x∂z=a and ∂y∂z=bInserting these into z=ax+by we havez=x∂x∂z+y∂y∂zwhich is equivalent to: x∂x∂z+y∂y∂z−z=02. Given z=ax then we will have the following: ∂x∂z=aInserting these into z=ax we havez=x∂x∂z
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