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Mathematics - Differentiation Question with Solution | TestHub

MathematicsDifferentiationDifferentiation of implicit functionsMedium2 minQB

MathematicsMediumsingle choice

Passage / Comprehension

Consider the functions defined implicitly by $y^{3} - 3 y + x = 0$ on various intervals of the real line. If $x \in (- \infty, - 2) \cup (2, \infty)$ , the equation defines a unique real-valued differentiable function $y = f (x)$ . If $x \in (- 2, 2)$ , it defines a unique real-valued differentiable function $y = g (x)$ satisfying $g (0) = 0$ . If $f (- 102) = 22$ ,

then $f^{''} (- 102) =$

Options:

Answer:

Solution:

Differentiate: $(3 y^{2} - 3) y^{'} + 1 = 0$ . At $x = - 102, y = 22, 3 y^{2} - 3 = 21$ , so $y^{'} = - \frac{1}{21}$ . Differentiate again: $6 y (y^{'})^{2} + (3 y^{2} - 3) y^{''} = 0$ , so $6 \cdot 22 \cdot \frac{1}{441} + 21 y^{''} = 0$ and $y^{''} = - \frac{12 2}{441 \cdot 21} = - \frac{12 2}{9261} = - \frac{4 2}{7 ^{3} \cdot 3 ^{2}}$ (since $9261 = 3^{3} \cdot 7^{3}$ ).

Stream:JEESubject:MathematicsTopic:DifferentiationSubtopic:Differentiation of implicit functions

⏱ 2mℹ️ Source: QB

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