We have, xy = ex-y⇒logxy = log ex-y⇒y log x = x-y log e⇒y log x = x - y⇒y log x + y = x⇒y =x1 + log xDifferentiating both sides with respect to x, we get dydx = 1 + log x×ddxx - x × ddx1+ log x1 + log x2⇒dydx = 1 + log x - x ×1x1 + log x2⇒dydx = log x1 + log x2

7 question n.o

We have, x^{y} = e^{x - y} \Rightarrow \log (x^{y}) = \log (e^{x - y}) \Rightarrow y \log x = (x - y) \log e \Rightarrow y \log x = x - y \Rightarrow y \log x + y = x \Rightarrow y = \frac{x}{1 + \log x} Differentiating both sides with respect to x, we get \frac{dy}{dx} = \frac{(1 + \log x) \times \frac{d}{dx} (x) - x \times \frac{d}{dx} (1 + \log x)}{{(1 + \log x)}^{2}} \Rightarrow \frac{dy}{dx} = \frac{1 + \log x - x \times \frac{1}{x}}{{(1 + \log x)}^{2}} \Rightarrow \frac{dy}{dx} = \frac{\log x}{{(1 + \log x)}^{2}}

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