Is there a formula for the second implicit derivative like for the first?

Question

The formula for the first implicit derivative dy/dx = -fx/fy where fx and fy are the respective partial derivatives. Is there such a formula for the second implicit derivative? (d2y/dx^2)

Answer 1

We know, that, in appropriate conditions, equation $F(x,y)=0$ defines function, called implicit function, $y=f(x)$.

Putting this function in equation gives identity $$F(x,f(x))=0$$ Differentiation of this identity gives $$F'_x(x,f(x))+F'_y(x,f(x))f'(x)=0$$ From where comes formula for first derivative. Now, differentiate it again, and putting expression for $f'$ we have $$f''_{x^2}=\frac{(F''_{xy}+F''_{y^2}f'_x)\cdot F'_x - (F''_{x^2}+F''_{xy}f'_x)\cdot F'_x}{F'^{2}_y}=\\ =\frac{2F'_x\cdot F'_y\cdot F''_{xy}-F'^{2}_y\cdot F''_{x^2}-F'^{2}_x\cdot F''_{y^2}}{F'^{3}_y}$$ Same way you can use for 3d and high derivatives. If/when you would like read it from book, then look at G.M. Fichtenholz, A Course of Differential and Integral Calculus, I volume.