Q.Find $\dfrac{dy}{dx}$ when $x$ and $y$ are connected by the relation: $\sec(x + y) = xy$.

Concept: Implicit Differentiation — differentiate both sides with respect to $x$, treating $y$ as a function of $x$, then solve for $\frac{dy}{dx}$.

Differentiate $\sec(x+y) = xy$:

Expand the left side:

Collect $\frac{dy}{dx}$ terms: …

The equation $\sec(x+y) = xy$ ties $x$ and $y$ together in a way that cannot be easily solved for $y$ in terms of $x$. That’s exactly when implicit differentiation shines: we treat $y$ as a function of $x$ and differentiate every term with respect to $x$, using the chain rule whenever we hit a $y$.

The left side is $\sec(x+y)$. Its derivative is $\sec(x+y)\tan(x+y)$, but because the argument is $x+y$, we must multiply by the derivative of $x+y$, which is $1 + \frac{dy}{dx}$.

The right side is $xy$, a product of $x$ and $y$. Using the product rule, its derivative is $1 \cdot y + x \cdot \frac{dy}{dx}$.

Now we set the derivatives equal and solve for $\frac{dy}{dx}$.

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

2. Left side: chain rule gives

3. Right side: product rule gives

4. So the equation becomes:

5. Expand the left side:

6. Bring terms with $\frac{dy}{dx}$ to one side, constants to the other:

7. Factor out $\frac{dy}{dx}$: …

Method: Implicit Differentiation of Equations with a Composite Trigonometric Term

Use this when one side of the equation is a trig function of a combination of $x$ and $y$, such as $\sec(x+y)$, $\sin(x-y)$, or $\cos(xy)$, equated to an algebraic expression in $x$ and $y$.

Steps

Step 1: Identify the inner argument that depends on both variables

If the trig function's argument is $x+y$ (or $x-y$, or a product like $xy$), note that differentiating it needs the chain rule, because the argument itself contains $y=y(x)$:

Step 2: Apply the chain rule to the trig term

For $\sec(x+y)$, use $\dfrac{d}{du}\sec u = \sec u\tan u$, then multiply by the derivative of the inner argument found in Step 1:

Step 3: Differentiate the other side using the product rule (if it's a product like $xy$)

$\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$ …

Common Mistakes

Mistake 1: Forgetting the chain-rule factor on the composite argument

Students often write $\frac{d}{dx}\sec(x+y)=\sec(x+y)\tan(x+y)$ and stop, without multiplying by $\frac{d}{dx}(x+y)=1+\frac{dy}{dx}$. Why it's wrong: $x+y$ is itself a function of $x$ (since $y$ depends on $x$), so the chain rule demands this extra factor. Correct approach: always write the derivative of $\sec(x+y)$ as $\sec(x+y)\tan(x+y)\cdot\left(1+\frac{dy}{dx}\right)$.

Mistake 2: Dropping the product rule on the right side

A common slip is differentiating $xy$ as just $y$ or just $x$, ignoring that both factors depend on $x$. Why it's wrong: $xy$ is a product, so its derivative is $y+x\frac{dy}{dx}$, not a single term. Correct approach: apply the product rule explicitly every time two $x$-dependent quantities are multiplied.

Mistake 3: Sign slip while collecting terms …