Q.Find $\dfrac{dy}{dx}$ when $x$ and $y$ are connected by the relation: $\tan^{-1}(x^2 + y^2) = a$.

Concept: Parametric/Implicit Differentiation — treat $y$ as a function of $x$ and differentiate both sides with respect to $x$.

Step 1: Differentiate both sides w.r.t. $x$:

Since $a$ is constant, RHS = 0.

Step 2: Using the chain rule:

Step 3: Compute the derivative inside: …

The equation $\tan^{-1}(x^2 + y^2) = a$ looks like it involves two variables, but notice that the right-hand side is a constant $a$. That means the entire expression inside the inverse tangent is fixed — $x^2 + y^2$ must be constant. So the curve is actually a circle centered at the origin. The derivative $\frac{dy}{dx}$ is the slope of the tangent to this circle, which we can find by implicit differentiation.

The key tool here is the derivative of $\tan^{-1} u$, which is $\frac{1}{1+u^2} \cdot \frac{du}{dx}$. Since $a$ is constant, its derivative is zero. Let’s go step by step.

1. Differentiate both sides with respect to $x$. The left side is $\tan^{-1}(x^2 + y^2)$. Using the chain rule:

The right side is $a$, a constant, so its derivative is $0$.

2. Compute $\frac{d}{dx}(x^2 + y^2)$. Differentiate term by term:

So:

3. Set up the equation. Putting it together:

4. Solve for $\frac{dy}{dx}$. The factor $\frac{1}{1 + (x^2 + y^2)^2}$ is never zero (it’s always positive), so we can multiply both sides by it without worry. This gives:

Divide through by 2: …

Method: Implicit Differentiation When the Equation Equals a Constant

Applies whenever the given relation sets a function of $x$ and $y$ equal to a constant (a letter like $a$, $c$, or $k$ that does not depend on $x$) rather than another expression in $x$.

Steps

Step 1: Recognise the constant and differentiate it to zero

Whatever function of $x,y$ sits on the equation's other side, the moment it equals a symbol representing a fixed number, its derivative with respect to $x$ is $0$:

This immediately simplifies the problem — you never need to compute a derivative for that side.

Step 2: Differentiate the function side using the chain rule

If the left side is, say, $\tan^{-1}(g(x,y))$, apply the standard inverse-trig derivative formula together with the chain rule on the inner expression:

where $\dfrac{dg}{dx}$ is found by differentiating $g(x,y)$ term by term, remembering the extra $\dfrac{dy}{dx}$ factor on every $y$-term.

Step 3: Set the differentiated expression equal to zero and simplify …

Common Mistakes

Mistake 1: Differentiating the constant $a$ as if it were a variable

Some students, seeing $\tan^{-1}(x^2+y^2)=a$, get confused and try to differentiate $a$ using some chain rule, or treat it as a function of $x$. Why it's wrong: $a$ is a fixed constant here, so its derivative with respect to $x$ is exactly $0$ — that's what makes the whole right side vanish. Correct approach: recognize immediately that a constant right-hand side means the derivative equation reduces to "(something) = 0".

Mistake 2: Forgetting the $\frac{dy}{dx}$ factor when differentiating $y^2$

Inside $x^2+y^2$, students often write $\frac{d}{dx}(y^2)=2y$ instead of $2y\frac{dy}{dx}$, treating $y$ like an independent variable rather than a function of $x$. Why it's wrong: $y$ is implicitly a function of $x$, so any $y$-term needs the chain rule. Correct approach: always attach $\frac{dy}{dx}$ whenever differentiating a $y$-term with respect to $x$.

Mistake 3: Getting stuck trying to isolate the arctan's derivative factor …