If `y= tan ^-1 ((x)/sqrt(a^2-x^2))` then `dy/dx =? `

Options

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

Answer

To find the derivative of the function \( y = \tan^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} \right) \), we can use the substitution method. Here’s the step-by-step solution:

### Step 1: Substitute \( x = a \sin \theta \)

Let \( x = a \sin \theta \). Then, we can express \( \sqrt{a^2 - x^2} \) as follows:

\[ \sqrt{a^2 - x^2} = \sqrt{a^2 - (a \sin \theta)^2} = \sqrt{a^2 (1 - \sin^2 \theta)} = a \cos \theta \]

### Step 2: Rewrite \( y \)

Substituting \( x \) and \( \sqrt{a^2 - x^2} \) into the equation for \( y \):

\[ y = \tan^{-1} \left( \frac{a \sin \theta}{a \cos \theta} \right) = \tan^{-1} (\tan \theta) \]

### Step 3: Simplify \( y \)

Since \( \tan^{-1} (\tan \theta) = \theta \), we have:

\[ y = \theta \]

### Step 4: Find \( \theta \) in terms of \( x \)

From our substitution \( x = a \sin \theta \), we can express \( \theta \) as:

\[ \theta = \sin^{-1} \left( \frac{x}{a} \right) \]

Thus, we can rewrite \( y \):

\[ y = \sin^{-1} \left( \frac{x}{a} \right) \]

### Step 5: Differentiate \( y \)

Now we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx} \left( \sin^{-1} \left( \frac{x}{a} \right) \right) \]

Using the derivative of the inverse sine function:

\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \left( \frac{x}{a} \right)^2}} \cdot \frac{d}{dx} \left( \frac{x}{a} \right) \]

### Step 6: Apply the chain rule

The derivative of \( \frac{x}{a} \) with respect to \( x \) is \( \frac{1}{a} \):

\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \frac{x^2}{a^2}}} \cdot \frac{1}{a} \]

### Step 7: Simplify the expression

Now, we simplify the expression:

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

### Final Answer

Thus, the derivative \( \frac{dy}{dx} \) is:

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