Найдите дифференциал функции y= y=3^{sin x/ln x}

Решите задачу:

$y=\frac{x^2+1}{x^3-1}\\\frac{dy}{dx}=\frac{(x^2+1)'(x^3-1)-(x^2+1)(x^3-1)'}{(x^3-1)^2}=\frac{2x(x^3-1)-3x^2(x^2+1)}{(x^3-1)^2}=\\=\frac{2x^4-2x-3x^4-3x^2}{(x^3-1)^2}=\frac{-x^4-3x^2-2x}{(x^3-1)^2}\\\\y=3^{\frac{\sin x}{\ln x}}\\\frac{dy}{dx}=3^{\frac{\sin x}{\ln x}}\ln3\cdot\left(\frac{\sin x}{\ln x}\right)'=3^{\frac{\sin x}{\ln x}}\ln3\cdot\frac{\cos x\ln x-\frac{\sin x} x}{\ln^2x}=\\=3^{\frac{\sin x}{\ln x}}\ln3\cdot\frac{x\cos x\ln x-{\sin x}}{x\ln^2x}$