Question

Differentiate \(\frac{\sin x}{\sqrt{\cos x}}\) \(\text{ with respect to } x\).

Hint:

Quick Tip: When differentiating a quotient, always use the quotient rule. Remember that the derivative of \( \cos x \) is \( -\sin x \), and the derivative of a square root function like \( \sqrt{u(x)} \) is \( \frac{1}{2\sqrt{u(x)}} \cdot u'(x) \).

Answer 1

We need to differentiate \( \frac{\sin x}{\sqrt{\cos x}} \) using the quotient rule. The quotient rule states that for a function \( \frac{u(x)}{v(x)} \), its derivative is given by: \[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2} \] Here, \( u(x) = \sin x \) and \( v(x) = \sqrt{\cos x} = (\cos x)^{1/2} \).
First, differentiate \( u(x) = \sin x \): \[ u'(x) = \cos x \] Next, differentiate \( v(x) = (\cos x)^{1/2} \): \[ v'(x) = \frac{1}{2} (\cos x)^{-1/2} \cdot (-\sin x) = -\frac{\sin x}{2 \sqrt{\cos x}} \] Now, applying the quotient rule: \[ \frac{d}{dx} \left( \frac{\sin x}{\sqrt{\cos x}} \right) = \frac{\sqrt{\cos x} \cdot \cos x - \sin x \cdot \left( -\frac{\sin x}{2 \sqrt{\cos x}} \right)}{(\sqrt{\cos x})^2} \] Simplifying: \[ = \frac{\cos x \sqrt{\cos x} + \frac{\sin^2 x}{2 \sqrt{\cos x}}}{\cos x} \] \[ = \frac{\sqrt{\cos x} (\cos^2 x + \frac{\sin^2 x}{2})}{\cos x} \] Hence, the derivative of \( \frac{\sin x}{\sqrt{\cos x}} \) is given by the above expression.