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Continuity and Differentiability

NCERT Class 12

CBSE

Grade 12

Question:

If $x^2 + y^2 = 100$, then $\dfrac{dy}{dx}$ at point $(6, 8)$ is:
(a) $-\dfrac{3}{4}$
(b) $\dfrac{3}{4}$
(c) $-\dfrac{4}{3}$
(d) $\dfrac{4}{3}$

Step-by-Step Solution

$y' = -x/y = -6/8 = -3/4$. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Evaluating derivative $= -3/4$: 1.0 Mark

Correct Answer: $-\dfrac{3}{4}$

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