Q.If $x\sin(a + y) + \sin a \cos(a + y) = 0$, prove that $\dfrac{dy}{dx} = \dfrac{\sin^2(a + y)}{\sin a}$.

Concept: Implicit Differentiation — differentiate both sides with respect to $x$, treating $y$ as a function of $x$.

Step 1: Differentiate the given equation term by term:

Step 2: Apply product rule to the first term:

Step 3: Collect terms with $\frac{dy}{dx}$:

Step 4: From the original equation, $x\sin(a+y) = -\sin a \cos(a+y)$, so $x = -\frac{\sin a \cos(a+y)}{\sin(a+y)}$. Substitute this into the coefficient: …

The problem asks us to prove a derivative from an implicit equation. The key is to see that $y$ is not isolated — it appears inside sine and cosine functions alongside $x$. This is a classic situation for implicit differentiation: differentiate every term with respect to $x$, treating $y$ as a function of $x$, and then solve for $\frac{dy}{dx}$.

Let’s first rewrite the given equation clearly:

Here $a$ is a constant (likely a parameter). Our goal: show that $\frac{dy}{dx} = \frac{\sin^2(a+y)}{\sin a}$.

1. Differentiate both sides with respect to $x$.

   Remember: $y$ depends on $x$, so $\frac{d}{dx} \sin(a+y) = \cos(a+y) \cdot \frac{dy}{dx}$, and similarly for $\cos(a+y)$.

   Differentiating term by term:

   • First term: $x \sin(a+y)$. Use the product rule:

• Second term: $\sin a \cos(a+y)$. Since $\sin a$ is constant:

The right-hand side is $0$, so its derivative is $0$.

Putting it together:

2. Collect the terms containing $\frac{dy}{dx}$. Move the term without $\frac{dy}{dx}$ to the other side:

Factor $\frac{dy}{dx}$:

3. Now we need to eliminate $x$ using the original equation. From the given: $x \sin(a+y) + \sin a \cos(a+y) = 0$, we can solve for $x$:

Substitute this $x$ into the bracket:

Simplify the first term:

Factor $\sin a$:

Combine inside the bracket over a common denominator $\sin(a+y)$: …

Method: Implicit Differentiation with Elimination Using the Original Equation

Use this method when, after implicit differentiation, the resulting expression for $\dfrac{dy}{dx}$ still contains a variable (often $x$) that you need to eliminate — the original equation itself usually supplies the substitution needed to clean it up.

Steps

Step 1: Differentiate the given equation term by term with respect to $x$

For a term like $x\sin(a+y)$, use the product rule together with the chain rule (since $a+y$ contains $y=y(x)$):

For a term like $\sin a\cos(a+y)$ (with $a$ constant), only the chain rule is needed:

Step 2: Collect $\dfrac{dy}{dx}$ terms and factor

Step 3: Solve the original equation for the leftover variable

The coefficient of $\dfrac{dy}{dx}$ still contains $x$, which needs to be removed. Go back to the original relation and solve it for $x$:

$x\sin(a+y) = -\sin a\cos(a+y) \quad\Rightarrow\quad x = -\frac{\sin a\cos(a+y)}{\sin(a+y)}$ …

Common Mistakes

Mistake 1: Dropping the product rule on $x\sin(a+y)$

Students sometimes differentiate $x\sin(a+y)$ as just $x\cos(a+y)\cdot\frac{dy}{dx}$, forgetting the extra $\sin(a+y)$ term that comes from differentiating the $x$ factor. Why it's wrong: $x\sin(a+y)$ is a product of $x$ and a $y$-dependent function, so the product rule contributes both $\sin(a+y)$ (from differentiating $x$) and $x\cos(a+y)\frac{dy}{dx}$ (from differentiating $\sin(a+y)$). Correct approach: always split a product like this into its two contributing terms before proceeding.

Mistake 2: Sign error differentiating $\cos(a+y)$

A very common slip is writing $\frac{d}{dx}\cos(a+y)=+\sin(a+y)\frac{dy}{dx}$ instead of $-\sin(a+y)\frac{dy}{dx}$. Why it's wrong: the derivative of cosine always carries a negative sign, and losing it flips the sign of the entire second term, corrupting the final collected coefficient. Correct approach: double-check the sign on every cosine derivative before combining terms. …