If \( 2^{x}+2^{y}=2^{x+y} \), then \( \frac{d y}{d x} \) is equal to \

1.	If \( 2^{x}+2^{y}=2^{x+y} \), then \( \frac{d y}{d x} \) is equal to \( \frac{2^{x}+2^{y}}{2^{x}-2^{y}} \) \( \frac{2^{x}+2^{y}}{1+2^{x+y}} \) \( 2^{x-y}\left[\frac{2^{y}-1}{1-2^{x}}\right] \) \( \frac{2^{x+y}-2^{x}}{2^{y}} \)
Answer» We have 2^x + 2^y = 2^{x + y} By differentiating both sides w.r.to x, we get 2^xln2 + 2^yln2 \(\frac{dy}{dx}=\) 2^x+y (1 + \(\frac{dy}{dx}\)) ln2 (\(\because\) \(\frac{d}{dx}a^x=a^x \) ln a) ⇒ ln 2 (2^x + 2^y\(\frac{dy}{dx}\)) = 2^x+y + 2^x+y \(\frac{dy}{dx}\) ⇒ (2^x+y - 2^y)\(\frac{dy}{dx}\) = 2^x - 2^x+y ⇒ 2^y(2^x - 1) \(\frac{dy}{dx}\) = 2^x(1 - 2^y) ⇒ \(\frac{dy}{dx}\) = \(\frac{2^x(1-2^y)}{2^y(2^x-1)}\) = \(\frac{2^{x-y}(1-2^y)}{2^x-1}\) = \(\frac{2^{x-y}(2^y-1)}{1-2^x}\)

If \( 2^{x}+2^{y}=2^{x+y} \), then \( \frac{d y}{d x} \) is equal to \( \frac{2^{x}+2^{y}}{2^{x}-2^{y}} \) \( \frac{2^{x}+2^{y}}{1+2^{x+y}} \) \( 2^{x-y}\left[\frac{2^{y}-1}{1-2^{x}}\right] \) \( \frac{2^{x+y}-2^{x}}{2^{y}} \)

Answer»

We have 2^x + 2^y = 2^{x + y}

By differentiating both sides w.r.to x, we get

2^xln2 + 2^yln2 \(\frac{dy}{dx}=\) 2^x+y (1 + \(\frac{dy}{dx}\)) ln2

(\(\because\) \(\frac{d}{dx}a^x=a^x \) ln a)

⇒ ln 2 (2^x + 2^y\(\frac{dy}{dx}\)) = 2^x+y + 2^x+y \(\frac{dy}{dx}\)

⇒ (2^x+y - 2^y)\(\frac{dy}{dx}\) = 2^x - 2^x+y

⇒ 2^y(2^x - 1) \(\frac{dy}{dx}\) = 2^x(1 - 2^y)

⇒ \(\frac{dy}{dx}\) = \(\frac{2^x(1-2^y)}{2^y(2^x-1)}\) = \(\frac{2^{x-y}(1-2^y)}{2^x-1}\)

= \(\frac{2^{x-y}(2^y-1)}{1-2^x}\)

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If \( 2^{x}+2^{y}=2^{x+y} \), then \( \frac{d y}{d x} \) is equal to \( \frac{2^{x}+2^{y}}{2^{x}-2^{y}} \) \( \frac{2^{x}+2^{y}}{1+2^{x+y}} \) \( 2^{x-y}\left[\frac{2^{y}-1}{1-2^{x}}\right] \) \( \frac{2^{x+y}-2^{x}}{2^{y}} \)

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