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Tiago Lima - Instagram: @professor_disciplinas_exatas • Seja a função , com , e , ƒ u, v,w = ln|uvw|( ) u = sen xz( ) v = x z2 w = 3x3 determine as derivadas parciais com relação a x, y e z da função. Resolução: , temos que: = ⋅ ⋅ ⋅ 𝜕ƒ u, v, w x ( ) 1 uvw 𝜕u 𝜕x 𝜕v 𝜕x 𝜕w 𝜕x = cos xz ⋅ z; = 2xz; = 8x 𝜕u 𝜕x ( ) 𝜕v 𝜕x 𝜕w 𝜕x 2 Assim, a derivada parcial em relação a x fica: = ⋅ cos xz ⋅ z ⋅ 2xz ⋅ 8x = 𝜕ƒ u, v, w x ( ) 1 sen xz ⋅ x z ⋅ 3x( ) 2 3 ( ) 2 16cos xz ⋅ z ⋅ x sen xz z ⋅ 3x ( ) 2 3 ( ) 5 = 𝜕ƒ u, v, w x ( ) 16 3 cos xz ⋅ z sen xz x ( ) ( ) 2 , temos que: = ⋅ ⋅ ⋅ 𝜕ƒ u, v, w y ( ) 1 uvw 𝜕u 𝜕y 𝜕v 𝜕y 𝜕w 𝜕y = 0; = 0; = 0 𝜕u 𝜕y 𝜕v 𝜕y 𝜕w 𝜕y Assim, a derivada parcial em relação a y fica: = ⋅ 0 ⋅ 0 ⋅ 0 𝜕ƒ u, v, w y ( ) 1 sen xz ⋅ x z ⋅ 3x( ) 2 3 = 0 𝜕ƒ u, v, w y ( ) , temos que: = ⋅ ⋅ ⋅ 𝜕ƒ u, v, w z ( ) 1 uvw 𝜕u 𝜕z 𝜕v 𝜕z 𝜕w 𝜕z = cos xz ⋅ x; = x ; = 0 𝜕u 𝜕z ( ) 𝜕v 𝜕z 2 𝜕w 𝜕z Assim, a derivada parcial em relação a z fica: = ⋅ cos xz ⋅ x ⋅ x ⋅ 0 𝜕ƒ u, v, w z ( ) 1 sen xz ⋅ x z ⋅ 3x( ) 2 3 ( ) 2 = 0 𝜕ƒ u, v, w z ( )
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