Re: [微積] 求dy/dx
: 如圖，想請問怎麼解出dy/dx?
: 先謝謝各位的回答！
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: Sent from JPTT on my LGE LG-K220.
exp(xy)+x^2-y^2=10
exp(xy)d(xy) +2xdx-2ydy=0 , let u=exp(xy)
u[xdy+ydx] +2xdx-2ydy=0
(2y-ux)dy=(2x+uy)dx
dy 2x+uy
── = ──── ,u代回可得解
dx 2y-ux
Hint:
1.Chain-rule : let u=f[g(x)] , du = f'[g(x)]g'(x)
2.let u=u(x,y) , d[exp(u)] = exp(u)du
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