Continuity and Differentiability: 3 marks Questions (CBSE Class 12)

3 different 3 marks questions on Continuity and Differentiability from CBSE Class 12 Maths board exams 2026, newest first.

If $(sin x)^{y} = y^{c o s x}$ , then find $\frac{d y}{d x}$ .

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Answer:

\frac{d y}{d x} = \frac{y ( y c o t x + s i n x l o g y )}{c o s x - y l o g ( s i n x )}

Take logs: $y lo g (sin x) = cos x lo g y$ .
Differentiate: $\frac{d y}{d x} lo g (sin x) + y cot x = - sin x lo g y + \frac{c o s x}{y} \frac{d y}{d x}$ .
$\frac{d y}{d x} (lo g (sin x) - \frac{c o s x}{y}) = - (y cot x + sin x lo g y)$ .
$\frac{d y}{d x} = \frac{y ( y c o t x + s i n x l o g y )}{c o s x - y l o g ( s i n x )}$ .

If $x y = e^{x - y}$ , then find $\frac{d y}{d x}$ .

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Answer:

\frac{d y}{d x} = \frac{y ( x - 1 )}{x ( y + 1 )}

Taking log: $lo g x + lo g y = x - y$ .
Differentiating: $\frac{1}{x} + \frac{1}{y} \frac{d y}{d x} = 1 - \frac{d y}{d x}$ .
$\frac{d y}{d x} (\frac{1}{y} + 1) = 1 - \frac{1}{x} \Rightarrow \frac{d y}{d x} \cdot \frac{1 + y}{y} = \frac{x - 1}{x}$ .
$\frac{d y}{d x} = \frac{y ( x - 1 )}{x ( y + 1 )}$ .

Differentiate $tan^{- 1} (\frac{1 + x ^{2} - 1 - x ^{2}}{1 + x ^{2} + 1 - x ^{2}})$ with respect to $cos^{- 1} x^{2}$ .

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Answer:

- \frac{1}{2}

Let $z = cos^{- 1} x^{2}$ , so $x^{2} = cos z$ with $0 \leq z \leq π$ , i.e. $0 \leq \frac{z}{2} \leq \frac{π}{2}$ .
$1 + cos z = 2 cos \frac{z}{2}$ , $1 - cos z = 2 sin \frac{z}{2}$ .
$y = tan^{- 1} (\frac{c o s \frac{z}{2} - s i n \frac{z}{2}}{c o s \frac{z}{2} + s i n \frac{z}{2}}) = tan^{- 1} (tan (\frac{π}{4} - \frac{z}{2})) = \frac{π}{4} - \frac{z}{2}$ , since $\frac{π}{4} - \frac{z}{2} \in [- \frac{π}{4}, \frac{π}{4}]$ .
$\frac{d y}{d z} = - \frac{1}{2}$ .

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