Differetiation

Question

U=log(x+2y+3z) then find Ux,Uy,Uz

Answer 1

\(U= log(x+2y+3z)\)

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»\(y=logx \)

»\( dy = {1\over x}\)

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\({dU\over dx} = {1\over (x+2y+3z)}.{d\over dx}(x+2y+3z)\)

⇒\({dU\over dx} = {1\over (x+2y+3z)}\) 

\({dU\over dy} = {1\over (x+2y+3z)}.{d\over dy}(x+2y+3z)\)

⇒\({dU\over dy} = {2\over (x+2y+3z)}\)

\({dU\over dz} = {1\over (x+2y+3z)}.{d\over dz}(x+2y+3z)\)

⇒\({dU\over dz} = {3\over (x+2y+3z)}\)

Comments

Thanks