If `y=sin^(-1)((4x)/(1+4x^(2)))` then what is `(dy)/(dx)` equal to ?

Options

A. A) `(1)/(1+4x^(2))`
B. B) `-(1)/(1+4x^(2))`
C. C) `(4)/(1+4x^(2))`
D. D) `(4x)/(1+4x^(2))`

Answer

To find the derivative \( \frac{dy}{dx} \) for the function \( y = \sin^{-1}\left(\frac{4x}{1 + 4x^2}\right) \), we will follow these steps:

### Step 1: Simplify the expression
We start with the function:
\[ y = \sin^{-1}\left(\frac{4x}{1 + 4x^2}\right) \]
We can recognize that the argument of the inverse sine function can be simplified using the double angle identity for sine.

### Step 2: Use a trigonometric substitution
Let:
\[ 2x = \tan(\theta) \]
Then:
\[ 4x = 2\tan(\theta) \]
And:
\[ 1 + 4x^2 = 1 + 4\tan^2(\theta) = 1 + \tan^2(\theta) \cdot 4 = \sec^2(\theta) \]
Thus, we can rewrite the expression:
\[ \frac{4x}{1 + 4x^2} = \frac{2\tan(\theta)}{\sec^2(\theta)} = 2\tan(\theta) \cos^2(\theta) = 2\sin(\theta) \]
This gives us:
\[ y = \sin^{-1}(2\sin(\theta)) \]

### Step 3: Simplify further
Using the identity \( \sin^{-1}(\sin(2\theta)) = 2\theta \):
\[ y = 2\theta \]
Now substituting back for \( \theta \):
\[ \theta = \tan^{-1}(2x) \implies y = 2\tan^{-1}(2x) \]

### Step 4: Differentiate \( y \) with respect to \( x \)
Now we differentiate \( y \):
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}(2x)) \]
Using the derivative of \( \tan^{-1}(u) \):
\[ \frac{d}{dx}(\tan^{-1}(u)) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = 2x \) and \( \frac{du}{dx} = 2 \):
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1 + (2x)^2} \cdot 2 = \frac{4}{1 + 4x^2} \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{4}{1 + 4x^2} \]

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