Question

If $y=\log \tan \left(\frac{x}{2}\right)+\sin ^{-1}(\cos x)$, then $\frac{d y}{d x}=$

• A. $\csc x$
• B. $\sin x + 1$
• C. $x$
• D. $\csc x - 1$

Hint:

The derivative of $\log\tan(x/2)$ is a highly useful standard result that evaluates directly to $\csc x$. Knowing this standard piece allows you to instantly combine it with the derivative of $-x$ (which is $-1$) to write down $\csc x - 1$ effortlessly!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for the first derivative $\frac{d y}{d x}$ of a composite function consisting of a logarithmic tangent term and an inverse trigonometric term.

Step 2: Key Formula or Approach:
We can differentiate the function term by term. To make differentiation straightforward, we can rewrite the inverse trigonometric part using the complementary angle identity: $$ \sin^{-1}(\cos x) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - x\right)\right) = \frac{\pi}{2} - x $$ The first term can be differentiated directly using the standard calculus chain rule.

Step 3: Detailed Explanation:
Let's differentiate the first term, $y_1 = \log \tan \left(\frac{x}{2}\right)$, using the chain rule: $$ \frac{d y_1}{d x} = \frac{1}{\tan\left(\frac{x}{2}\right)} \cdot \frac{d}{dx}\left(\tan\left(\frac{x}{2}\right)\right) $$ $$ \frac{d y_1}{d x} = \frac{1}{\tan\left(\frac{x}{2}\right)} \cdot \sec^2\left(\frac{x}{2}\right) \cdot \frac{1}{2} $$ Converting these functions into their fundamental sine and cosine expressions: $$ \frac{d y_1}{d x} = \frac{\cos\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \cdot \frac{1}{\cos^2\left(\frac{x}{2}\right)} \cdot \frac{1}{2} = \frac{1}{2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)} $$ Applying the sine double-angle identity $2\sin\theta\cos\theta = \sin(2\theta)$: $$ \frac{d y_1}{d x} = \frac{1}{\sin x} = \csc x $$ Now, let's differentiate the second simplified term, $y_2 = \frac{\pi}{2} - x$: $$ \frac{d y_2}{d x} = 0 - 1 = -1 $$ Combining both derivatives together gives the complete result: $$ \frac{d y}{d x} = \csc x - 1 $$

Step 4: Final Answer:
The derivative of the function is $\frac{d y}{d x} = \csc x - 1$, corresponding to option (D).