Limits, Continuity & Differentiability
Chain Rule and Composite Functions
Grade 12
Question:
<p>Let <span class='math'>f(x) = e^x</span>, <span class='math'>g(x) = \sin^{-1} x</span> and <span class='math'>h(x) = f[g(x)]</span>, then <span class='math'>\frac{h'(x)}{h(x)}</span> is equal to</p>
<p>(a) <span class='math'>\frac{\sin^{-1} x}{1-x^2}</span></p>
<p>(b) <span class='math'>\frac{1}{\sqrt{1-x^2}}</span></p>
<p>(c) <span class='math'>\sin^{-1} x</span></p>
<p>(d) <span class='math'>\frac{1}{\sqrt{1-x^2}}</span></p>
Step-by-Step Solution
Key Concept: Apply chain rule to composite function and simplify the ratio of derivative to function.
<p><strong>Step 1:</strong> Find <span class='math'>h(x) = f[g(x)] = e^{\sin^{-1}x}</span></p><p><strong>Step 2:</strong> Differentiate using chain rule: <span class='math'>h'(x) = e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1-x^2}}</span></p><p><strong>Step 3:</strong> <span class='math'>\frac{h'(x)}{h(x)} = \frac{e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1-x^2}}}{e^{\sin^{-1}x}} = \frac{1}{\sqrt{1-x^2}}</span></p>
Correct Answer: B