If `y=tan^(-1)((5-2tansqrt(x))/(2+5tan sqrt(x)))` then what is `(dy)/(dx)` equal to ?

Options

A. A) `-(1)/(2sqrt(x))`
B. B) `1`
C. C) `-1`
D. D) `(1)/(2sqrt(x))`

Answer

To find the derivative \( \frac{dy}{dx} \) for the function

\[ y = \tan^{-1}\left(\frac{5 - 2\tan(\sqrt{x})}{2 + 5\tan(\sqrt{x})}\right), \]

we can follow these steps:

### Step 1: Rewrite the function
We can simplify the expression inside the inverse tangent function. Notice that we can factor out a 2 from both the numerator and the denominator:

\[ y = \tan^{-1}\left(\frac{2\left(\frac{5}{2} - \tan(\sqrt{x})\right)}{2\left(1 + \frac{5}{2}\tan(\sqrt{x})\right)}\right). \]

This simplifies to:

\[ y = \tan^{-1}\left(\frac{5/2 - \tan(\sqrt{x})}{1 + (5/2)\tan(\sqrt{x})}\right). \]

### Step 2: Apply the formula for inverse tangent
Using the formula for the inverse tangent of a difference:

\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right), \]

we can let \( a = \frac{5}{2} \) and \( b = \tan(\sqrt{x}) \). Thus, we can express \( y \) as:

\[ y = \tan^{-1}\left(\frac{5}{2}\right) - \tan^{-1}(\tan(\sqrt{x})). \]

Since \( \tan^{-1}(\tan(\theta)) = \theta \) for \( \theta \) in the principal range, we have:

\[ y = \tan^{-1}\left(\frac{5}{2}\right) - \sqrt{x}. \]

### Step 3: Differentiate with respect to \( x \)
Now we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = 0 - \frac{d}{dx}(\sqrt{x}). \]

The derivative of \( \sqrt{x} \) is:

\[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}. \]

Thus, we have:

\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}. \]

### Final Answer
Therefore, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}. \]

---