Question

Differentiate \(\frac{\sin x}{\sqrt{\cos x}}\) 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 will differentiate \( \frac{\sin x}{\sqrt{\cos x}} \) using the quotient rule.

The quotient rule for a function \( \frac{u(x)}{v(x)} \) is:

\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2} \]

In this case, \( u(x) = \sin x \) and \( v(x) = \sqrt{\cos x} = (\cos x)^{1/2} \).

Step 1: Find the derivative of \( u(x) \):

\[ u'(x) = \cos x \]

Step 2: Find the derivative of \( v(x) \):

\[ v'(x) = \frac{1}{2} (\cos x)^{-1/2} \cdot (-\sin x) = -\frac{\sin x}{2 \sqrt{\cos x}} \]

Step 3: Apply the quotient rule formula:

\[ \frac{d}{dx} \left( \frac{\sin x}{\sqrt{\cos x}} \right) = \frac{\sqrt{\cos x} \cdot \cos x - \sin x \cdot \left( -\frac{\sin x}{2 \sqrt{\cos x}} \right)}{(\sqrt{\cos x})^2} \]

Step 4: Simplify the resulting expression:

\[ = \frac{\cos x \sqrt{\cos x} + \frac{\sin^2 x}{2 \sqrt{\cos x}}}{\cos x} \]

Final Answer:

\[ \frac{\sqrt{\cos x} \left( \cos^2 x + \frac{\sin^2 x}{2} \right)}{\cos x} \]

The derivative of \( \frac{\sin x}{\sqrt{\cos x}} \) is the expression derived above.