A) \[\frac{1}{1+{{a}^{{{\cos }^{-1}}x}}}\]
B) \[\frac{1}{{{(1+{{a}^{{{\cos }^{-1}}x}})}^{2}}}\]
C) \[\frac{-1}{1+{{a}^{{{\cos }^{-1}}x}}}\]
D) none of these
Correct Answer: D
Solution :
We have, \[y=\frac{{{a}^{{{\cos }^{-1}}x}}}{1+{{a}^{{{\cos }^{-1}}x}}},z={{a}^{{{\cos }^{-1}}x}}\] \[\Rightarrow \] \[y=\frac{z}{1+z}\] On differentiating w.r.t. z, we get \[\frac{dy}{dz}=\frac{(1+z)-z(1)}{{{(1+z)}^{2}}}\] \[=\frac{1}{{{(1+z)}^{2}}}\] \[=\frac{1}{{{(1+{{a}^{{{\cos }^{-1}}x}})}^{2}}}\] Now, \[z={{a}^{{{\cos }^{-1}}x}}\] On differentiating w.r.t. x, we get \[\frac{dz}{dx}={{a}^{{{\cos }^{-1}}x}}\log a\left( -\frac{1}{\sqrt{1-{{x}^{2}}}} \right)\] \[\therefore \] \[\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}\] \[=\frac{1}{{{(1+{{a}^{{{\cos }^{-1}}x}})}^{2}}}.{{a}^{{{\cos }^{-1}}x}}\left( -\frac{1}{\sqrt{1-{{x}^{2}}}} \right)\log a\] \[=-\frac{{{a}^{{{\cos }^{-1}}x}}}{{{(1+{{a}^{{{\cos }^{-1}}x}})}^{2}}}.\frac{1}{\sqrt{1-{{x}^{2}}}}.\log \,a\]
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