Question

derivative of e^(x-y)=x^y prove that derivative is logx /(log e*x)^2

Answer 1

Step-by-step explanation:

Answer 2

Answer:

${x}^{y } = {e}^{x - y} \\ log( {x}^{y} ) = log( {e}^{x - y} ) \\ y \: log(x) = (x - y) log(e) \: \: \: \: \: \: \: \: \:( log \: e = 1) \\ \: \: \: y \: log(x) = x - y \\ y \: log(x) + y = x \\ y(log(x) + 1) = x \\ y = \frac{x}{log(x) + 1} \\ \frac{dy}{dx} = \frac{(log(x) + 1)(1) - x( \frac{1}{x} + 0) }{ {(1 + log(x))}^{2} } \\ \frac{dy}{dx} = \frac{log(x) + 1 - 1}{ {(1 + log(x))}^{2} } \\ \frac{dy}{dx} = \frac{log(x)}{ {(1 + log(x))}^{2} } \\ \frac{dy}{dx } = \frac{log(x)}{ {(log(e) +log(x)) }^{2} } \: \: \: \: \: \: \: \: \: \: \: (log \: e = 1) \\ \frac{dy}{dx} = \frac{log(x)}{ {log(ex)}^{2} } \: \: \: \: \: \: \: \: (log \: a + log \: b = log(ab)) \\ hence \: the \: proof$

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