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Mathematics Question on Differentiation

Supposef(x)=(2x+2−x)tanxtan−1(x2−x+1)(7x2+3x+1)3f(x)=\frac{(2^x+2^{-x})tanx\sqrt{tan^{-1}(x^2-x+1)}}{(7x^2+3x+1)^{3}} then the value of f′(0)f'(0) is equal to

A

π\pi

B

0

C

π\sqrt\pi

D

π2\frac{\pi}{2}

Answer

π\sqrt\pi

Explanation

Solution

Step 1: Calculate f′(0)f'(0) Using the Definition of Derivative

f′(0)=lim⁡h→0f(h)−f(0)hf'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}

Step 2: Evaluate f(h)−f(0)f(h) - f(0)

Substitute x=hx = h and x=0x = 0 into f(x)f(x):

f′(0)=lim⁡h→0(2h+2−h)tan⁡htan⁡−1(h2−h+1)−0(7h2+3h+1)3⋅hf'(0) = \lim_{h \to 0} \frac{\left(2^h + 2^{-h}\right) \tan h \sqrt{\tan^{-1}(h^2 - h + 1)} - 0}{(7h^2 + 3h + 1)^3 \cdot h}

Step 3: Simplify the Expression

Using the limit properties, we get:

f′(0)=πf'(0) = \sqrt{\pi}

So, the correct answer is: π\sqrt{\pi}