04 - Determine as derivadas parciais de primeira e segunda ordem das seguintes funções:
(h) f(alpha, beta)=(2 alpha+3 beta)^202

Passo 1

O enunciado fornece a função
fcolon R^2 to R, f(alpha,beta) = (2alpha+3beta)^202
Precisamos calcular as derivadas parciais de primeira ordem partial f/partial alpha, partial f/partial beta e as de segunda ordem partial^2 f/partial alpha^2, partial^2 f/partial alphapartial beta e partial^2 f/partial beta^2. Vamos usar a regra da cadeia, pois f é composta de u=2alpha+3beta elevada a 202.

Passo 2

Para partial fpartial alpha, defina u = 2alpha+3beta. Então f(u)=u^202 e
partial fpartial alpha = 202,u^201,partial upartial alpha = 202,(2alpha+3beta)^201cdot2
partial fpartial alpha = 404,(2alpha+3beta)^201

Passo 3

Para partial fpartial beta, de modo análogo,
partial fpartial beta = 202,(2alpha+3beta)^201,partial (2alpha+3beta)partial beta = 202,(2alpha+3beta)^201cdot3
partial fpartial beta = 606,(2alpha+3beta)^201

Passo 4

Para partial^2 fpartial alpha^2, derive partial f/partial alpha = 404,(2alpha+3beta)^201 em relação a alpha e use partial(2alpha+3beta)/partial alpha=2:
partial^2 fpartial alpha^2 = 404cdot 201,(2alpha+3beta)^200cdot2 = 162408,(2alpha+3beta)^200

Passo 5

Para o derivado misto partial^2 fpartial beta,partial alpha, derive partial f/partial alpha em relação a beta usando partial(2alpha+3beta)/partial beta=3:
partial^2 fpartial beta,partial alpha = 404cdot 201,(2alpha+3beta)^200cdot3 = 243612,(2alpha+3beta)^200

Passo 6

Para partial^2 fpartial beta^2, derive partial f/partial beta = 606,(2alpha+3beta)^201 em relação a beta, com partial(2alpha+3beta)/partial beta=3:
partial^2 fpartial beta^2 = 606cdot 201,(2alpha+3beta)^200cdot3 = 365418,(2alpha+3beta)^200

Resposta

partial fpartial alpha = 404,(2alpha+3beta)^201, partial fpartial beta = 606,(2alpha+3beta)^201,
partial^2 fpartial alpha^2 = 162408,(2alpha+3beta)^200, partial^2 fpartial beta,partial alpha = 243612,(2alpha+3beta)^200, partial^2 fpartial beta^2 = 365418,(2alpha+3beta)^200

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