Q.If $x = \sqrt{a^{\sin^{-1} t}}, y = \sqrt{a^{\cos^{-1} t}}$, show that $\frac{dy}{dx} = -\frac{y}{x}$

Concept: Implicit Differentiation — we simplify the given parametric forms using exponent rules, then differentiate.

First, rewrite the expressions using powers:

Take natural logs to simplify differentiation:

Differentiate both with respect to $t$: …

The key here is to recognize that $x$ and $y$ are both functions of $t$, and we need $\frac{dy}{dx}$, not $\frac{dy}{dt}$ or $\frac{dx}{dt}$ individually. The standard parametric approach is to compute $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, but the expressions involve inverse trigonometric functions in the exponents. That’s where implicit differentiation shines — it lets us avoid messy exponent manipulation by working directly with the relationships.

Let’s rewrite the given equations for clarity:

We want to show $\frac{dy}{dx} = -\frac{y}{x}$. Notice that if we multiply $x$ and $y$, something interesting happens:

a constant! This suggests $x$ and $y$ are inversely related, but let’s prove it step by step.

1. Differentiate $x$ with respect to $t$. Since $x = a^{\frac{1}{2} \sin^{-1} t}$, take the natural log: $\log x = \frac{1}{2} \sin^{-1} t \cdot \log a$. Differentiate both sides with respect to $t$:

So,

2. Differentiate $y$ with respect to $t$. Similarly, $\log y = \frac{1}{2} \cos^{-1} t \cdot \log a$. Differentiate:

(Recall $\frac{d}{dt} \cos^{-1} t = -\frac{1}{\sqrt{1 - t^2}}$.)

Thus,

3. Form the ratio $\frac{dy}{dx}$. …

Method: Parametric Differentiation with a Variable Base (Logarithmic Route)

This method applies whenever a curve is given through a third variable (a parameter — commonly $t$ or $\theta$) instead of $y$ written directly as a function of $x$.

Steps

Step 1: Recognise the parametric form

If you are given $x = f(\text{param})$ and $y = g(\text{param})$ instead of $y = h(x)$, do not try to eliminate the parameter first — it is often messy or impossible. Differentiate each equation separately with respect to the parameter instead.

Step 2: Differentiate $x$ and $y$ with respect to the parameter

Use the ordinary rules (product rule, chain rule, standard derivatives) to find $\dfrac{dx}{d\theta}$ (or $\dfrac{dx}{dt}$) and $\dfrac{dy}{d\theta}$ (or $\dfrac{dy}{dt}$).

Step 3: Divide — the parametric-derivative formula

This is justified by the chain rule: $\dfrac{dy}{d\theta} = \dfrac{dy}{dx}\cdot\dfrac{dx}{d\theta}$, so dividing recovers $\dfrac{dy}{dx}$.

Step 4: Simplify with trigonometric identities where possible …

Common Mistakes

Mistake 1: Dropping the negative sign in $\frac{d}{dt}\cos^{-1}t=-\frac{1}{\sqrt{1-t^2}}$.

Why it's wrong: $\sin^{-1}t$ and $\cos^{-1}t$ have derivatives that are exact negatives of each other, and mixing up the sign here directly flips the sign of the final answer, turning $-\frac{y}{x}$ into $+\frac{y}{x}$. Correct approach: keep the two standard inverse-trig derivatives written side by side and apply the correct one to each variable.

Mistake 2: Trying to isolate $x$ and $y$ individually as functions of $t$ before dividing. …