What is \(\frac{{dy}}{{dx}}\) at x = 1 equal to?

Concept:

\({\log _a}{x^b} = b \times \;{\log _a}x\;and\;{\log _a}a = 1\)

\(\frac{{d\left( {u\;v} \right)}}{{dx}} = v \times \frac{{du}}{{dx}} + u \times \frac{{dv}}{{dx}}\)

\(\frac{{d\;\left( {\ln x} \right)}}{{dx}} = \frac{1}{x}\)

Calculation:

Given: x^y = e^x-y

By applying ln both the sides we get

⇒ y × ln x = ln (e^x-y) = x – y

⇒ y × ln x = x – y

⇒ x = y × (1 + ln x)      ------(1)

Now by differentiating both the sides with respect to x we get,

\( \Rightarrow 1 = \left( {1 + \ln x} \right) \times \frac{{dy}}{{dx}} + \frac{y}{x}\)

\( \Rightarrow \frac{{dy}}{{dx}} = \frac{{1 - \frac{y}{x}}}{{1 + \ln x}}\)

Now by using equation (1), we get

\( \Rightarrow \frac{{dy}}{{dx}} = \frac{{\ln x}}{{{{\left( {1 + \ln x} \right)}^2}}}\)

So, \(\frac{{dy}}{{dx}}\;at\;x = 1\;is\;0\)