Chain rule for $$\frac{d}{dx}\left(\frac{-16x}{2y}\right)$$

Expert intro: This problem is about differentiating an expression that contains both $x$ and a function $y(x)$. The key idea is to treat $y$ as dependent on $x$ and apply the chain rule correctly.

Detailed walkthrough

Let $f(x)=\frac{-16x}{2y}=-8x\,y^{-1}.$ Since $y$ depends on $x$, differentiate using the product rule and chain rule:

$\frac{d}{dx}[-8x\,y^{-1}] = -8\frac{d}{dx}(x)y^{-1} + (-8x)\frac{d}{dx}(y^{-1}).$

Now compute each derivative:

$\frac{d}{dx}(x)=1,$

and

$\frac{d}{dx}(y^{-1})=-y^{-2}\frac{dy}{dx}.$

So

= -8y^{-1} + (-8x)(-y^{-2})\frac{dy}{dx}.$$ That simplifies to $$\frac{d}{dx}\left(\frac{-16x}{2y}\right) = -\frac{8}{y}+\frac{8x}{y^2}\frac{dy}{dx}.$$ If you are checking the work shown in the prompt, the missing idea is that differentiating $y^{-1}$ requires the chain rule, so a factor of $\frac{dy}{dx}$ must appear. ### 💡 Pitfall guide A common mistake is to treat $y$ as a constant and differentiate only the $x$ term. That is not valid if $y=y(x)$. Another frequent error is writing the derivative of $y^{-1}$ as just $-y^{-2}$ and forgetting the extra factor $\frac{dy}{dx}$. ### 🔄 Real-world variant If $y$ were actually a constant, then the derivative would be much simpler: $$\frac{d}{dx}\left(\frac{-16x}{2y}\right)=\frac{-16}{2y}=-\frac{8}{y}.$$ But when $y$ depends on $x$, the full chain rule result is $$-\frac{8}{y}+\frac{8x}{y^2}\frac{dy}{dx}.$$ ### 🔍 Related terms chain rule, implicit differentiation, product rule