Differentiation Question 437
Question: If $ 2^{x}+2^{y}={2^{x+y}} $ , then $ \frac{dy}{dx}= $
[MP PET 1995; AMU 2000]
Options:
A) $ {2^{x-y}}\frac{2^{y}-1}{2^{x}-1} $
B) $ {2^{x-y}}\frac{2^{y}-1}{1-2^{x}} $
C) $ \frac{2^{x}+2^{y}}{2^{x}-2^{y}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
On differentiating $ 2^{x}\log 2+2^{y}\log 2.\frac{dy}{dx} $
$ =2^{x}{{.2}^{y}}\frac{dy}{dx}.\log 2+2^{y}{{.2}^{x}}\log 2 $
Therefore $ 2^{x}+2^{y}\frac{dy}{dx}={2^{x+y}}\frac{dy}{dx}+{2^{x+y}} $
Therefore $ \frac{dy}{dx}=\frac{{2^{x+y}}-2^{x}}{2^{y}-{2^{x+y}}}={2^{x-y}}\frac{2^{y}-1}{1-2^{x}} $ .
BETA
