If y = ^{-1} ≤ft( {{x} - x}{1 + x³/2}} ), then y'(1) is equal to

Correct Answer is : −14-1/4−41

Correct Answer is : −14-1/4−41

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BITSAT 2024 Shift 1 Paper

Question : 91 of 130

Marks: +1, -0

y = \tan^{-1} \left( \frac{\sqrt{x} - x}{1 + x^{3/2}} \right)

y'(1)

Solution:

Given the function

y = \tan^{-1} \left( \frac{\sqrt{x} - x}{1 + x^{3/2}} \right)

We can rewrite this expression using an identity for inverse tangent:

y = \tan^{-1} \left( \frac{\sqrt{x} - x}{1 + \sqrt{x} \cdot x} \right)

This can be expressed as:

y = \tan^{-1}(\sqrt{x}) - \tan^{-1}(x)

This transformation uses the identity:

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

To find the derivative

y'

with respect to

x

, we use the differentiation of inverse tangent functions:

y' = \frac{d}{dx} \left[ \tan^{-1}(\sqrt{x}) - \tan^{-1}(x) \right]

The derivative is calculated as follows:

y' = \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{1}{2 \sqrt{x}} - \frac{1}{1 + x^2}

Now, evaluate

y'(1)

y'(1) = \frac{1}{1 + 1} \cdot \frac{1}{2 \cdot 1} - \frac{1}{1 + 1} = \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}

Thus, the value of

y'(1)

-\frac{1}{4}

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