Suppose $$f(x) = \frac{(2^x + 2^{-x})\tan x \sqrt{\tan^{-1}(x^2 - x + 1)}}{(7x^2 + 3x + 1)^3}$$. Then the value of $$f'(0)$$ is equal to

 we use the definition of the derivative at $$x = 0$$: $$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}.$$ First, evaluate $$f(0)$$:

• Numerator: $$(2^0 + 2^{-0}) \tan 0 \sqrt{\tan^{-1}(0^2 - 0 + 1)} = (1 + 1) \cdot 0 \cdot \sqrt{\tan^{-1}(1)} = 2 \cdot 0 \cdot \sqrt{\pi/4} = 0$$, since $$\tan^{-1}(1) = \pi/4$$.
• Denominator: $$(7(0)^2 + 3(0) + 1)^3 = 1^3 = 1$$.

Thus, $$f(0) = 0/1 = 0$$. So,

$$f'(0) = \lim_{h \to 0} \frac{f(h)}{h} = \lim_{h \to 0} \frac{1}{h} \cdot \frac{(2^h + 2^{-h}) \tan h \sqrt{\tan^{-1}(h^2 - h + 1)}}{(7h^2 + 3h + 1)^3}.$$

$$\frac{f(h)}{h} = (2^h + 2^{-h}) \cdot \left( \frac{\tan h}{h} \right) \cdot \sqrt{\tan^{-1}(h^2 - h + 1)} \cdot \frac{1}{(7h^2 + 3h + 1)^3}.$$

Now, evaluate the limit as $$h \to 0$$ for each part:

• $$\lim_{h \to 0} (2^h + 2^{-h}) = 2^0 + 2^0 = 1 + 1 = 2$$.
• $$\lim_{h \to 0} \frac{\tan h}{h} = 1$$, since $$\tan h \sim h$$ near 0.
• $$\lim_{h \to 0} \sqrt{\tan^{-1}(h^2 - h + 1)} = \sqrt{\tan^{-1}(1)} = \sqrt{\pi/4} = \sqrt{\pi}/2$$, as $$\tan^{-1}(1) = \pi/4$$.
• $$\lim_{h \to 0} \frac{1}{(7h^2 + 3h + 1)^3} = \frac{1}{(1)^3} = 1$$, since the denominator approaches 1.

$$f'(0) = 2 \cdot 1 \cdot \frac{\sqrt{\pi}}{2} \cdot 1 = \sqrt{\pi}.$$

The function and its components are well-behaved at $$x = 0$$:

• $$2^x + 2^{-x}$$ is continuous and differentiable.
• $$\tan x$$ is continuous and differentiable near 0 (as $$\cos x \neq 0$$).
• $$\tan^{-1}(x^2 - x + 1)$$ is continuous and differentiable since $$x^2 - x + 1 \geq 3/4 > 0$$ for all $$x$$, and its square root is defined and differentiable.
• The denominator $$(7x^2 + 3x + 1)^3$$ is never zero (minimum value at $$x = -3/14$$ is positive).

Thus, the limit exists and equals $$\sqrt{\pi}$$.