KCET · Maths · Differentiation
If \(2^{x}+2^{y}=2^{x+y}\), then \(\frac{d y}{d x}\) is
- A \(2^{y-x}\)
- B \(-2^{y-x}\)
- C \(2^{x-y}\)
- D \(\frac{2^{y}-1}{2^{x}-1}\)
Answer & Solution
Correct Answer
(B) \(-2^{y-x}\)
Step-by-step Solution
Detailed explanation
We have,
\(2^{x}+2^{y}=2^{x+y}...(i)\)
On differentiating Eq. (i) w.r.t. \(x\), we get
\(2^{x} \log 2+2^{y} \log 2 \frac{d y}{d x}\)
\(=2^{x+y} \log 2\left(1+\frac{d y}{d x}\right)\)
\(\Rightarrow \quad 2^{x}+2^{y} \frac{d y}{d x}=2^{x+y}\left(1+\frac{d y}{d x}\right)\)
\(\Rightarrow \quad 2^{x}-2^{x+y}=\frac{d y}{d x}\left(2^{x+y}-2^{y}\right)\)
\(\Rightarrow \quad-2^{y}=\frac{d y}{d x}\left(2^{x}\right)\)
\(\Rightarrow \quad \frac{d y}{d x}=-2^{y} / 2^{x}=-2^{y-x}\)
\(2^{x}+2^{y}=2^{x+y}...(i)\)
On differentiating Eq. (i) w.r.t. \(x\), we get
\(2^{x} \log 2+2^{y} \log 2 \frac{d y}{d x}\)
\(=2^{x+y} \log 2\left(1+\frac{d y}{d x}\right)\)
\(\Rightarrow \quad 2^{x}+2^{y} \frac{d y}{d x}=2^{x+y}\left(1+\frac{d y}{d x}\right)\)
\(\Rightarrow \quad 2^{x}-2^{x+y}=\frac{d y}{d x}\left(2^{x+y}-2^{y}\right)\)
\(\Rightarrow \quad-2^{y}=\frac{d y}{d x}\left(2^{x}\right)\)
\(\Rightarrow \quad \frac{d y}{d x}=-2^{y} / 2^{x}=-2^{y-x}\)
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