Observe the following statements:I. f(x) = a | vedclass

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(A) For statement $I$:$f(x) = a x^{41} + b x^{-40}$$f^{\prime}(x) = 41 a x^{40} - 40 b x^{-41}$$f^{\prime \prime}(x) = 41 	imes 40 a x^{39} + 40 	im

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$I$ is true, but $II$ is false

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(A) For statement $I$:$f(x) = a x^{41} + b x^{-40}$$f^{\prime}(x) = 41 a x^{40} - 40 b x^{-41}$$f^{\prime \prime}(x) = 41 	imes 40 a x^{39} + 40 	imes 41 b x^{-42} = 1640 a x^{39} + 1640 b x^{-42}$$\frac{f^{\prime \prime}(x)}{f(x)} = \frac{1640(a x^{39} + b x^{-42})}{a x^{41} + b x^{-40}} = \frac{1640(a x^{39} + b x^{-42})}{x^2(a x^{39} + b x^{-42})} = 1640 x^{-2}$.Thus, statement $I$ is true.For statement $II$:Let $y = 	an ^{-1}\left(\frac{2 x}{1-x^2}ight)$. Using the substitution $x = 	an 	heta$, we get $y = 	an ^{-1}(	an 2 	heta) = 2 	heta = 2 	an ^{-1} x$.Then $\frac{d y}{d x} = \frac{2}{1+x^2}$.Since the given statement claims the derivative is $\frac{1}{1+x^2}$, statement $II$ is false.Therefore, $I$ is true, but $II$ is false.

Observe the following statements:
$I. f(x) = a x^{41} + b x^{-40} \Rightarrow \frac{f^{\prime \prime}(x)}{f(x)} = 1640 x^{-2}$
$II. \frac{d}{d x} \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) = \frac{1}{1+x^2}$
Which of the following is correct?

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Observe the following statements:$I. f(x) = a x^{41} + b x^{-40} \Rightarrow \frac{f^{\prime \prime}(x)}{f(x)} = 1640 x^{-2}$$II. \frac{d}{d x} \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) = \frac{1}{1+x^2}$Which of the following is correct?