Utilize a Regra da Cadeia para determinar as derivadas parciais
partial upartial alpha, partial upartial beta, partial upartial gamma
sabendo que
u=ln (x) tan (y) e^t raizx y, x=alpha^2 beta, y=beta^2 gamma, t=gamma^2 alpha

Passo 1

O enunciado pede para calcular as derivadas parciais partial upartial alpha, partial upartial beta, partial upartial gamma da função
u = \ln(x),\tan(y),e^{t\sqrt{xy}}
sujeita às mudanças de variáveis
x = \alpha^2\beta,
y = \beta^2\gamma,
t = \gamma^2\alpha.
O conteúdo principal é a regra da cadeia em várias variáveis, e vamos precisar de
  1. ux, uy, ut, 2) as derivadas de x,y,t em relação a alpha,beta,gamma, e 3) combinar tudo.

Passo 2

Para achar ux, uy e ut, trate u= f(x)g(y)h(t,x,y) com
f(x)=\ln(x),
g(y)=\tan(y),
h= e^{t\sqrt{xy}}.
Então
partial upartial x = f'(x),g(y),h + f(x),g(y),partial hpartial x = 1xtan(y)e^traizxy + ln(x)tan(y)e^traizxycdott,y2raizxy
partial upartial y = f(x)g'(y)h + f(x)g(y)partial hpartial y = ln(x)sec^2(y)e^traizxy + ln(x)tan(y)e^traizxycdott,x2raizxy
partial upartial t = f(x)g(y),hcdotpartialpartial t(traizxy) = ln(x)tan(y)e^traizxyraizxy

Passo 3

Agora calcule as derivadas de
x=\alpha^2\beta,\quad y=\beta^2\gamma,\quad t=\gamma^2\alpha
em relação a alpha,beta,gamma:
partial xpartial alpha=2alphabeta, partial xpartial beta=alpha^2, partial xpartial gamma=0;
partial ypartial alpha=0, partial ypartial beta=2betagamma, partial ypartial gamma=beta^2;
partial tpartial alpha=gamma^2, partial tpartial beta=0, partial tpartial gamma=2gammaalpha

Passo 4

Use agora a regra da cadeia:
partial upartial alpha = ux,xalpha + uy,yalpha + ut,talpha, partial upartial beta = ux,xbeta + uy,ybeta + ut,tbeta, partial upartial gamma = ux,xgamma + uy,ygamma + ut,tgamma
Substituindo ux,uy,ut e as derivadas de x,y,t, obtemos as expressões finais das três derivadas parciais.

Resposta

As derivadas parciais finais são:
partial upartial alpha = tan(y)e^traizxy gl[2alphabetax + alphabetaln(x),t,yraizxy + ln(x)raizxy,gamma^2gr],
partial upartial beta = tan(y)e^traizxy gl[alpha^2x + alpha^2ln(x),t,y2raizxygr] + 2betagamma,ln(x)e^traizxy gl[sec^2(y) + tan(y),t,x2raizxygr],
partial upartial gamma = ln(x)e^traizxy gl[beta^2sec^2(y) + beta^2tan(y),t,x2raizxy + 2gammaalpha,tan(y)raizxygr]

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