Question

If \( y = \frac{x^2 + 3x + 4}{e^x \cos x} \), find \( \frac{dy}{dx} \).

Hint:

Use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{v u' - u v'}{v^2} \) for differentiation.

Answer 1

Step 1: Use quotient rule. \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \] where \( u = x^2 + 3x + 4 \) and \( v = e^x \cos x \). 

Step 2: Differentiate numerator and denominator. \[ \frac{du}{dx} = 2x + 3, \quad \frac{dv}{dx} = e^x \cos x - e^x \sin x \] 

Step 3: Apply the quotient rule. \[ \frac{dy}{dx} = \frac{(2x+3)e^x \cos x - (x^2+3x+4)(e^x \cos x - e^x \sin x)}{(e^x \cos x)^2} \]