Problem text:
Show that
$$ x^{y} + \sin(y) = 1 $$
Defines $y$ as a function of $x$ in the surrounding region of $(1, 0)$ and find $y'(x)$.
Textbook answer:
$$y'(x) = \frac{-yx^{y-1}}{x^{y} \ln(x) + \cos y}$$
How do I find $y'(x)$? The problem appears in a chapter about Jacobian determinants and the implicit function theorem, so perhaps they are of use?
Just use implicit derivation knowing that is allowed by the implicit function theorem: $$x^{y} + \sin(y) = 1\implies x^y(y'\log x+\frac{y}{x})+y'\cos y=0\implies y'(\cos y+x^y\log x)+yx^{y-1}=0\implies \\y'=\frac{-yx^{y-1}}{\cos y+x^{y}\log x}$$