MCQ
If ${2^x} + {2^y} = {2^{x + y}}$, then ${{dy} \over {dx}} = $
- A${2^{x - y}}{{{2^y} - 1} \over {{2^x} - 1}}$
- ✓${2^{x - y}}{{{2^y} - 1} \over {1 - {2^x}}}$
- C${{{2^x} + {2^y}} \over {{2^x} - {2^y}}}$
- DNone of these
✓
Answer
Correct option: B.
${2^{x - y}}{{{2^y} - 1} \over {1 - {2^x}}}$
b
(b) On differentiating ${2^x}\log 2 + {2^y}\log 2.\frac{{dy}}{{dx}}$
(b) 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$
==> ${2^x} + {2^y}\frac{{dy}}{{dx}} = {2^{x + y}}\frac{{dy}}{{dx}} + {2^{x + y}}$
==> $\frac{{dy}}{{dx}} = \frac{{{2^{x + y}} - {2^x}}}{{{2^y} - {2^{x + y}}}} = {2^{x - y}}\frac{{{2^y} - 1}}{{1 - {2^x}}}$.
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