If \[y=\frac{{{a}^{{{\cos }^{-1}}x}}}{1+{{a}^{{{\cos }^{-1}}x}}}\]and \[z={{a}^{{{\cos }^{-1}}x}}\], then \[\frac{dy}{dx}\]= [MP PET 1994]

\[\frac{1}{1+{{a}^{{{\cos }^{-1}}x}}}\]

\[-\frac{1}{1+{{a}^{{{\cos }^{-1}}x}}}\]

\[\frac{1}{{{(1+{{a}^{{{\cos }^{-1}}x}})}^{2}}}\]

If \[y=\frac{{{a}^{{{\cos }^{-1}}x}}}{1+{{a}^{{{\cos }^{-1}}x}}}\]and

1.	If \[y=\frac{{{a}^{{{\cos }^{-1}}x}}}{1+{{a}^{{{\cos }^{-1}}x}}}\]and \[z={{a}^{{{\cos }^{-1}}x}}\], then \[\frac{dy}{dx}\]= [MP PET 1994]
A.	\[\frac{1}{1+{{a}^{{{\cos }^{-1}}x}}}\]
B.	\[-\frac{1}{1+{{a}^{{{\cos }^{-1}}x}}}\]
C.	\[\frac{1}{{{(1+{{a}^{{{\cos }^{-1}}x}})}^{2}}}\]
D.	None of these
Answer» D. None of these

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If \[y=\frac{{{a}^{{{\cos }^{-1}}x}}}{1+{{a}^{{{\cos }^{-1}}x}}}\]and \[z={{a}^{{{\cos }^{-1}}x}}\], then \[\frac{dy}{dx}\]= [MP PET 1994]

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