Question

If $y = \log(\sin x) + \log(\cos x)$, then $\frac{dy}{dx}$ is:

• A. \( \cos x - \sin x \)
• B. \( \sin x - \cos x \)
• C. \( \cot x - \tan x \)
• D. \( -\sin x - \cos x \)

Hint:

Using logarithmic properties simplifies the equation first: \(y = \log(\sin x \cdot \cos x)\). Differentiating gives \(\frac{1}{\sin x \cos x} \cdot (\cos^2 x - \sin^2 x) = \frac{\cos^2 x}{\sin x \cos x} - \frac{\sin^2 x}{\sin x \cos x} = \cot x - \tan x\).

Answer 1

The Correct Option is C

Concept: We can determine the derivative either by differentiating each term directly using the Chain Rule: \[ \frac{d}{dx}(\log(f(x))) = \frac{1}{f(x)} \cdot f'(x) \] Alternatively, we can first apply logarithmic properties to condense the expression before executing the differentiation step: \[ \log a + \log b = \log(a \cdot b) \]

Step 1: Differentiating the terms individually using the Chain Rule.
Given function: \[ y = \log(\sin x) + \log(\cos x) \] Differentiating both sides with respect to \(x\): \[ \frac{dy}{dx} = \frac{d}{dx}[\log(\sin x)] + \frac{d}{dx}[\log(\cos x)] \] Applying the chain rule onto both components: \[ \frac{dy}{dx} = \left(\frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x)\right) + \left(\frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x)\right) \] Since \(\frac{d}{dx}(\sin x) = \cos x\) and \(\frac{d}{dx}(\cos x) = -\sin x\): \[ \frac{dy}{dx} = \left(\frac{1}{\sin x} \cdot \cos x\right) + \left(\frac{1}{\cos x} \cdot (-\sin x)\right) \]

Step 2: Simplifying the expression using trigonometric ratios.
Using basic definitions \(\frac{\cos x}{\sin x} = \cot x\) and \(\frac{\sin x}{\cos x} = \tan x\): \[ \frac{dy}{dx} = \cot x - \tan x \]