Class CBSE Class 12 Mathematics Derivatives Q #1372
KNOWLEDGE BASED
REMEMBER
5 Marks 2025 AISSCE(Board Exam) LA
If $x=a\left(\cos\theta+\log\tan\frac{\theta}{2}\right)$ and $y=\sin\theta$, then find $\frac{d^{2}y}{dx^{2}}$ at $\theta=\frac{\pi}{4}$.
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Detailed Solution<\/h3>\r\n <\/div>\r\n\r\n
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Step 1: Find $\frac{dx}{d\theta}$

Given $x=a\left(\cos\theta+\log\tan\frac{\theta}{2}\right)$, differentiate with respect to $\theta$:\r\n$$ \frac{dx}{d\theta} = a\left(-\sin\theta + \frac{1}{\tan\frac{\theta}{2}} \cdot \sec^2\frac{\theta}{2} \cdot \frac{1}{2}\right) $$\r\n$$ \frac{dx}{d\theta} = a\left(-\sin\theta + \frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}} \cdot \frac{1}{\cos^2\frac{\theta}{2}} \cdot \frac{1}{2}\right) $$\r\n$$ \frac{dx}{d\theta} = a\left(-\sin\theta + \frac{1}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right) $$\r\n$$ \frac{dx}{d\theta} = a\left(-\sin\theta + \frac{1}{\sin\theta}\right) $$\r\n$$ \frac{dx}{d\theta} = a\left(\frac{1-\sin^2\theta}{\sin\theta}\right) = a\frac{\cos^2\theta}{\sin\theta} $$

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Step 2: Find $\frac{dy}{d\theta}$

Given $y = a\sin\theta$, differentiate with respect to $\theta$:\r\n$$ \frac{dy}{d\theta} = a\cos\theta $$

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Step 3: Find $\frac{dy}{dx}$

Using the chain rule, $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$:\r\n$$ \frac{dy}{dx} = \frac{a\cos\theta}{a\frac{\cos^2\theta}{\sin\theta}} = \frac{\sin\theta}{\cos\theta} = \tan\theta $$

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Step 4: Find $\frac{d^2y}{dx^2}$

Differentiate $\frac{dy}{dx} = \tan\theta$ with respect to $x$:\r\n$$ \frac{d^2y}{dx^2} = \frac{d}{dx}(\tan\theta) = \sec^2\theta \cdot \frac{d\theta}{dx} $$\r\nSince $\frac{dx}{d\theta} = a\frac{\cos^2\theta}{\sin\theta}$, then $\frac{d\theta}{dx} = \frac{\sin\theta}{a\cos^2\theta}$.\r\n$$ \frac{d^2y}{dx^2} = \sec^2\theta \cdot \frac{\sin\theta}{a\cos^2\theta} = \frac{1}{\cos^2\theta} \cdot \frac{\sin\theta}{a\cos^2\theta} = \frac{\sin\theta}{a\cos^4\theta} $$

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Step 5: Evaluate $\frac{d^2y}{dx^2}$ at $\theta = \frac{\pi}{4}$

Substitute $\theta = \frac{\pi}{4}$ into the expression for $\frac{d^2y}{dx^2}$:\r\n$$ \frac{d^2y}{dx^2}\Big|_{\theta=\frac{\pi}{4}} = \frac{\sin\frac{\pi}{4}}{a\cos^4\frac{\pi}{4}} = \frac{\frac{1}{\sqrt{2}}}{a\left(\frac{1}{\sqrt{2}}\right)^4} = \frac{\frac{1}{\sqrt{2}}}{a\left(\frac{1}{4}\right)} = \frac{4}{a\sqrt{2}} = \frac{4\sqrt{2}}{2a} = \frac{2\sqrt{2}}{a} $$

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\r\n Final Answer: $\frac{2\sqrt{2}}{a}$<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>

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\r\n Pedagogical Audit<\/span>\r\n <\/div>\r\n
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\r\n Bloom's Analysis:<\/span> \r\n This is an REMEMBER<\/span> question because it requires recalling differentiation rules and trigonometric identities to solve the problem.\r\n <\/div>\r\n
\r\n Knowledge Dimension:<\/span> \r\n PROCEDURAL<\/span>\r\n
\r\n Justification:<\/span> \r\n The question requires applying differentiation techniques and trigonometric identities in a specific sequence to arrive at the solution.<\/span>\r\n <\/div>\r\n <\/div>\r\n
\r\n Syllabus Audit:<\/span> \r\n In the context of CBSE Class 12, this is classified as KNOWLEDGE<\/span>. It directly tests the student's ability to apply differentiation rules and trigonometric identities, which are core concepts in the syllabus.<\/span>. [NATURE_REASONING]\r\n <\/div>\r\n <\/div>\r\n <\/div>

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