Q.For the curve $\sqrt{x} + \sqrt{y} = 1$, $\dfrac{dy}{dx}$ at $\left(\dfrac{1}{4}, \dfrac{1}{4}\right)$ is __________.

Concept: Implicit differentiation of a curve with fractional powers.

Step 1 – Differentiate implicitly

$\sqrt{x} + \sqrt{y} = 1$

Write as $x^{1/2} + y^{1/2} = 1$. Differentiating both sides with respect to $x$:

$\frac{1}{2}x^{-1/2} + \frac{1}{2}y^{-1/2} \cdot \frac{dy}{dx} = 0$

Step 2 – Solve for $\frac{dy}{dx}$

Multiply through by $2$:

$x^{-1/2} + y^{-1/2} \cdot \frac{dy}{dx} = 0$ …

Why this approach works

The equation $\sqrt{x} + \sqrt{y} = 1$ is not written as $y = f(x)$ — it’s an implicit relation between $x$ and $y$. To find $\frac{dy}{dx}$, we differentiate both sides with respect to $x$, treating $y$ as a function of $x$. This is implicit differentiation, and it works perfectly even when solving for $y$ explicitly would be messy.

A key point: $\sqrt{x}$ and $\sqrt{y}$ are only defined for $x \ge 0$, $y \ge 0$, and the curve is smooth (differentiable) at interior points like $\left(\frac14,\frac14\right)$ because both $x$ and $y$ are positive there — no corner or cusp issues.

Step-by-step solution

1. Write the given equation clearly.

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

Remember: $\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$, and for $\sqrt{y}$, we use the chain rule: $\frac{d}{dx}(\sqrt{y}) = \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx}$.

So:

3. Solve for $\frac{dy}{dx}$.

Multiply through by $2$ to simplify:

Isolate the derivative term:

Multiply both sides by $\sqrt{y}$:

$\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} = -\sqrt{\frac{y}{x}}$ …

Method: Implicit Differentiation for Equations with Radical or Fractional-Power Terms

This method solves "find $\dfrac{dy}{dx}$" problems for curves given by an implicit equation involving square roots or fractional powers (such as $\sqrt{x}+\sqrt{y}=c$), where isolating $y$ explicitly is impractical.

Steps

Step 1: Rewrite radicals as fractional powers

Convert every square root to the form $(\cdot)^{1/2}$ so the power rule applies mechanically: $\sqrt{x}=x^{1/2}$, $\sqrt{y}=y^{1/2}$.

Step 2: Differentiate both sides with respect to $x$, treating $y$ as $y(x)$

Apply the power rule to each term. Every $x$-term differentiates normally; every $y$-term needs the chain rule, which multiplies in a factor of $\dfrac{dy}{dx}$:

Step 3: Collect and solve for $\dfrac{dy}{dx}$ …

Common Mistakes

Mistake 1: Forgetting the chain-rule factor on $\sqrt{y}$

A student differentiates $\sqrt{y}$ as $\frac{1}{2\sqrt{y}}$ and stops there, treating $y$ as if it were the independent variable $x$. Why it's wrong: $y$ is itself a function of $x$, so by the chain rule $\frac{d}{dx}(\sqrt y)=\frac{1}{2\sqrt y}\cdot\frac{dy}{dx}$ — dropping the trailing $\frac{dy}{dx}$ silently sets it equal to $1$, which is not valid implicit differentiation. Correct approach: multiply every derivative of a $y$-term by $\frac{dy}{dx}$.

Mistake 2: Losing the negative sign while isolating $\frac{dy}{dx}$

After differentiating to get $\frac{1}{2\sqrt x}+\frac{1}{2\sqrt y}\cdot\frac{dy}{dx}=0$, students sometimes drop the minus sign when solving, especially since substituting $x=y=\frac14$ makes both sides look symmetric and "cancel-able." Why it's wrong: it produces $+1$ instead of the correct $-1$, and misses that the curve is genuinely decreasing at that point. Correct approach: move the $x$-term to the other side carefully, keeping its sign, before dividing. …