Q.Find $\frac{dy}{dx}$ in the following: $\sin^2 x + \cos^2 y = 1$

Differentiate $\sin^2 x+\cos^2 y=1$ implicitly:

which is $\sin 2x-\sin 2y\,\dfrac{dy}{dx}=0$. Hence …

We differentiate $\sin^2 x+\cos^2 y=1$ with respect to $x$, treating $y$ as a function of $x$ and using the chain rule on each squared trig term.

Differentiate each term

and the right side, a constant, differentiates to $0$.

Assemble and solve

$\sin 2x-\sin 2y\,\frac{dy}{dx}=0\;\Rightarrow\;\frac{dy}{dx}=\frac{\sin 2x}{\sin 2y}.$ …

Method: Recognising When an Identity-Looking Equation Still Needs Implicit Differentiation

Some equations (like $\sin^2 x+\cos^2 y=1$) look like the familiar Pythagorean identity $\sin^2\theta+\cos^2\theta=1$, but here the two trig functions use different variables, $x$ and $y$ — so it is a genuine relationship between two variables, not an automatic identity, and it must be differentiated implicitly.

Steps

Step 1: Confirm the two variables are genuinely different

If the same letter appeared in both trig functions (e.g. $\sin^2 x + \cos^2 x$), the expression would be identically $1$ for every $x$, and differentiating would trivially give $0=0$. Here $y$ is a separate, dependent variable, so the equation instead defines a curve relating $x$ and $y$, and $\frac{dy}{dx}$ is a real, meaningful slope.

Step 2: Differentiate the $x$-term normally

Step 3: Differentiate the $y$-term with the chain rule, keeping the sign

$\frac{d}{dx}\big(\cos^2 y\big) = 2\cos y\cdot(-\sin y)\cdot\frac{dy}{dx} = -2\sin y\cos y\,\frac{dy}{dx}.$ …

Common Mistakes

Mistake 1: Assuming the equation is a trivial identity because it "looks like" $\sin^2\theta+\cos^2\theta=1$

Since $\sin^2 x+\cos^2 y=1$ resembles the Pythagorean identity, it's tempting to think both sides are automatically equal for all $x$ and treat the equation as always true, missing that $\frac{dy}{dx}$ is genuinely asked for. Why it's wrong: the identity only holds when the SAME angle appears in both terms; here $x$ and $y$ are different variables, so the equation defines an actual curve with a real slope. Correct approach: read the variables carefully before assuming a familiar-looking equation is an identity — differentiate implicitly as usual.

Mistake 2: Dropping the negative sign from differentiating $\cos y$ …