Q.Differentiate $\dfrac{x}{\sin x}$ w.r.t. $\sin x$.

"Differentiate $f$ with respect to $g$" means find $\dfrac{df/dx}{dg/dx}$. Let $y=\dfrac{x}{\sin x}$ and $u=\sin x$.

Quotient rule:

and $\dfrac{du}{dx}=\cos x$.

So …

To differentiate one function with respect to another (not with respect to $x$), use the chain-rule ratio

where here $y=\dfrac{x}{\sin x}$ and $u=\sin x$.

1. Differentiate $y$ with respect to $x$

$y=\dfrac{x}{\sin x}$ is a quotient, so by the quotient rule

2. Differentiate $u=\sin x$

3. Divide …

Method: Differentiating One Function with Respect to Another (Not with Respect to $x$)

Use this whenever a question asks you to differentiate $f(x)$ with respect to $g(x)$, rather than with respect to $x$ itself — the answer is not $\dfrac{df}{dx}$, it is a genuinely different ratio.

Steps

Step 1: Set up both functions as functions of $x$

Let $y = f(x)$ (the function you're asked to differentiate) and $u = g(x)$ (the function you're differentiating with respect to).

Step 2: Differentiate each one with respect to $x$ separately

Find $\dfrac{dy}{dx}$ using whatever rule the expression for $y$ needs (Quotient Rule, Product Rule, etc.), and find $\dfrac{du}{dx}$ similarly for $u$.

Step 3: Form the ratio

$\frac{dy}{du} = \frac{dy/dx}{du/dx}, \qquad \text{provided } \frac{du}{dx} \ne 0.$ …

Common Mistakes

Mistake 1: Reading "differentiate w.r.t. $\sin x$" as "differentiate w.r.t. $x$"

Why it's wrong: computing $\dfrac{dy}{dx}$ via the quotient rule and stopping there answers a different question — the problem asks for the rate of change of $y=\tfrac{x}{\sin x}$ relative to $u=\sin x$, not relative to $x$. Correct approach: after finding $dy/dx$, also find $du/dx=\cos x$ and divide: $\dfrac{dy}{du}=\dfrac{dy/dx}{du/dx}$.

Mistake 2: Sign error in the quotient rule on $\dfrac{x}{\sin x}$ …