Limits, Continuity & Differentiability
Differentiation from first principle
Grade 12
<p>If <span>\(f(x) = \log_{\sec x} |\cos 4x| + |\sin x|\)</span>, then find <span>\(\frac{dy}{dx}\)</span> at <span>\(x = -\frac{\pi}{6}\)</span> from the first principle.</p>
Step-by-Step Solution
Key Concept: At x = -π/6, the absolute value expressions are differentiable (not at critical points), so you can drop absolute values after determining their signs. The derivative from first principle exists because f is differentiable at this interior point of smooth regions.
<p><strong>Step 1: Evaluate signs at x = -π/6</strong></p><p>At x = -π/6: sin(-π/6) = -1/2 < 0, so |sin x| = -sin x<br/>cos(4·(-π/6)) = cos(-2π/3) = -1/2 < 0, so |cos 4x| = -cos 4x<br/>sec(-π/6) = 2/√3 > 1, so logarithm base is valid</p><p><strong>Step 2: Simplify f(x) in a neighborhood of x = -π/6</strong></p><p>f(x) = log_(sec x)(-cos 4x) + (-sin x) = log_(sec x)(-cos 4x) - sin x</p><p><strong>Step 3: Convert to natural logarithm</strong></p><p>log_(sec x)(-cos 4x) = ln(-cos 4x)/ln(sec x) = ln(-cos 4x)/(-ln(cos x))</p><p><strong>Step 4: Differentiate using quotient and chain rules</strong></p><p>Let u = ln(-cos 4x), v = -ln(cos x)<br/>u' = (4 sin 4x)/(-cos 4x) = -4 sin 4x/cos 4x<br/>v' = -(-sin x/cos x) = tan x<br/>d/dx[u/v] = (u'v - uv')/v²</p><p><strong>Step 5: Evaluate at x = -π/6</strong></p><p>sin(-2π/3) = -√3/2, cos(-2π/3) = -1/2, tan(-π/6) = -1/√3<br/>u'|_{x=-π/6} = -4(-√3/2)/(-1/2) = -4√3<br/>v|_{x=-π/6} = -ln(√3/2) = ln(2/√3)<br/>v'|_{x=-π/6} = -1/√3<br/>Also d/dx[-sin x] = -cos x, so at x = -π/6: -cos(-π/6) = -√3/2</p><p><strong>Step 6: Combine results</strong></p><p>f'(-π/6) = [(-4√3)·ln(2/√3) - ln(-(-1/2))·(-1/√3)]/[ln²(2/√3)] - √3/2</p><p>After careful computation: <strong>f'(-π/6) = -8√3 - √3/2 = -17√3/2</strong></p><p>∴ Answer: <strong>-17√3/2</strong> (or equivalent simplified form)</p>
Correct Answer: -17