If y= ^{-1}≤ft({1-4^{x}}{1+4^{x}}), then d y/d x at x=1 is

Correct Answer is : 4log⁡254 2/554log2

Correct Answer is : 4log⁡254 2/554log2

Test Index

MHTCET 16 Apr 2026 Shift 2 Paper

Question : 123 of 150

Marks: +1, -0

y=\cos ^{-1}\left(\frac{1-4^{x}}{1+4^{x}}\right)

\frac{d y}{d x}

x=1

Solution:

Concept:

Simplify the inverse cosine expression using the half-angle identity, then differentiate a standard inverse tangent function.

Explanation:

Given:

y=\cos^{-1}\left(\frac{1-4^{x}}{1+4^{x}}\right)

So,

\cos y=\frac{1-4^{x}}{1+4^{x}}

Use the identity:

\tan^{2}\frac{y}{2}=\frac{1-\cos y}{1+\cos y}

Substitute

\cos y

\tan^{2}\frac{y}{2}=\frac{1-\frac{1-4^{x}}{1+4^{x}}}{1+\frac{1-4^{x}}{1+4^{x}}}

Simplify:

\tan^{2}\frac{y}{2}=\frac{\frac{2\cdot 4^{x}}{1+4^{x}}}{\frac{2}{1+4^{x}}}=4^{x}

Since

y\in[0,\pi]

, we have

\frac{y}{2}\in\left[0,\frac{\pi}{2}\right]

, so

\tan\frac{y}{2}

is positive:

\tan\frac{y}{2}=2^{x}

Thus,

\frac{y}{2}=\tan^{-1}(2^{x})

y=2\tan^{-1}(2^{x})

Differentiate with respect to

x

\frac{dy}{dx}=2\cdot\frac{1}{1+(2^{x})^{2}}\cdot 2^{x}\log 2

=\frac{2\cdot 2^{x}\log 2}{1+4^{x}}

x=1

\frac{dy}{dx}=\frac{2\cdot 2\log 2}{1+4}=\frac{4\log 2}{5}

Answer:

\frac{4\log 2}{5}

So, the correct option is A.

Go to Question:

Prev Question

Next Question