Ordinary Derivative of a Function of a Polynomial

Question

I want to find higher order derivatives of: $$ f(x)=\frac{1}{4}(x^2-1)^2+K; \quad\quad K=\text{constant} $$ However, I want to find the derivatives with respect to the variable $y$ where we are given the relationship that $x=2y-1$. My confusion is that following two approaches I am getting slightly different answers. Approach $1$ was to substitute from the start and then calculate the derivatives which gave (after expanding): $$ f(2y-1) = 4y^4-8y^3+4y^2+K $$ Differentiating and applying the chain rule to the RHS, gives: $$ \frac{df}{dy}(2)=16y^3-24y^2+8y \implies \frac{df}{dy}=8y^3-12y^2+4y $$ My other approach was to calculate the derivatives with respect to $x$ and then use substitution afterwards: $$ f(x)=\frac{1}{4}x^4-\frac{1}{2}x^2+\frac{1}{4} +K $$ $$ \frac{df}{dx} = x^3-x $$ Using the chain rule, we have $\frac{df}{dx}=\frac{df}{dy}\frac{dy}{dx}=\frac{df}{dy}\cdot\frac{1}{2}$ . But this would imply: $$ \frac{df}{dy} = 2\frac{df}{dx}=2(x^3-x)=2[(2y-1)^3-(2y-1)]=16y^3-24y^2+8y $$ I believe I should get the same thing with either approach but I can't tell why they disagree by a factor of $2$. For reference, what I'm actually interested in is the second and third derivatives of this function. Help would be much appreciated in resolving my mistake.

Answer 1

The second one is right. The factor of $2$ on the left side in first approach is not there. When you view $f$ as function of $y$ you get $4y^{4}-8y^{3}+4y^{2}+K$ and you just differentiate this w.r.t $y$. Chain rule is not involved here.