Differentiation Question 132
Question: 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]
Options:
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
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=\frac{{a^{{{\cos }^{-1}}x}}}{1+{a^{{{\cos }^{-1}}x}}},z={a^{{{\cos }^{-1}}x}} $
$ \Rightarrow y=\frac{z}{1+z} $
$ \Rightarrow \frac{dy}{dz}=\frac{(1+z)1-z(1)}{{{(1+z)}^{2}}}=\frac{1}{{{(1+z)}^{2}}} $
$ =\frac{1}{{{(1+{a^{{{\cos }^{-1}}x}})}^{2}}} $ .
BETA
