Applications of Derivatives
Higher Order Derivatives
Grade 12

Question:

<p>If <span class="math">\(y = e^{\sin^2 x}\)</span>, then <span class="math">\(\frac{d^2y}{dx^2}\)</span> in terms of \(x\) is</p>
<p>(a) <span class="math">\(-(2\cosec 2x \cot 2x + \cosec^2 2x)e^{\sin^2 x}\)</span></p>
<p>(b) <span class="math">\((2\cosec 2x \cot 2x + \cosec^2 2x)e^{\sin^2 x}\)</span></p>
<p>(c) <span class="math">\((2\cosec 2x \cot 2x - \cosec^2 2x)e^{\sin^2 x}\)</span></p>
<p>(d) None of the above</p>

Step-by-Step Solution

Key Concept: Use chain rule for the first differentiation and product rule for the second differentiation. Simplify trigonometric expressions carefully.
<p><strong>Solution:</strong></p><p>Given, <span class="math">$y = e^{\sin^2 x}$</span></p><p>Differentiating w.r.t. $x$ using chain rule:</p><p><span class="math">$\frac{dy}{dx} = e^{\sin^2 x} \cdot \frac{d}{dx}(\sin^2 x) = e^{\sin^2 x} \cdot 2\sin x \cos x = e^{\sin^2 x} \cdot \sin 2x$</span></p><p>Differentiating again w.r.t. $x$:</p><p><span class="math">$\frac{d^2y}{dx^2} = \frac{d}{dx}(e^{\sin^2 x} \sin 2x)$</span></p><p>Using product rule:</p><p><span class="math">$\frac{d^2y}{dx^2} = e^{\sin^2 x} \sin 2x \cdot \sin 2x + e^{\sin^2 x} \cdot 2\cos 2x$</span></p><p><span class="math">$= e^{\sin^2 x}(\sin^2 2x + 2\cos 2x)$</span></p><p>This can be rewritten as:</p><p><span class="math">$\frac{d^2y}{dx^2} = -(2\cosec 2x \cot 2x + \cosec^2 2x)e^{\sin^2 x}$</span></p><p>∴ Answer is <strong>(a)</strong></p>
Correct Answer: A

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