Applications of Derivatives
Exponential and Trigonometric Functions
Grade 12
Question:
<p>The functions \(u = e^x \sin x\) and \(v = e^x \cos x\) satisfy the equation</p>
<p>(a) \(v\frac{du}{dx} - u\frac{dv}{dx} = u^2 + v^2\)</p>
<p>(b) \(v\,du + u\,dv = u^2 + v^2\)</p>
<p>(c) \(\frac{du}{dx} + \frac{dv}{dx} = 2v\)</p>
<p>(d) \(\frac{du}{dx} + \frac{dv}{dx} = 2u\)</p>
Step-by-Step Solution
Key Concept: Use product rule for derivatives of exponential times trigonometric functions, then verify the algebraic relations between the derivatives and original functions.
<p><strong>Solution:</strong></p><p>Given: $u = e^x \sin x$, $v = e^x \cos x$</p><p>Differentiating both:</p><p>$\frac{du}{dx} = e^x(\sin x + \cos x)$ ... (i)</p><p>$\frac{dv}{dx} = e^x(\cos x - \sin x)$ ... (ii)</p><p><strong>Checking option (a):</strong></p><p>$v\frac{du}{dx} - u\frac{dv}{dx} = e^x\cos x \cdot e^x(\sin x + \cos x) - e^x\sin x \cdot e^x(\cos x - \sin x)$</p><p>$= e^{2x}[\cos x(\sin x + \cos x) - \sin x(\cos x - \sin x)]$</p><p>$= e^{2x}[\cos x \sin x + \cos^2 x - \sin x \cos x + \sin^2 x]$</p><p>$= e^{2x}(\cos^2 x + \sin^2 x) = e^{2x}$</p><p>Also, $u^2 + v^2 = e^{2x}\sin^2 x + e^{2x}\cos^2 x = e^{2x}$ ✓</p><p><strong>Checking option (c):</strong></p><p>$\frac{du}{dx} + \frac{dv}{dx} = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x\cos x = 2v$ ✓</p>
Correct Answer: A, C