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Chapter 6: Problem 78

If \(e^{y}=y^{x}\), prove that \(\frac{d y}{d x}=\frac{(\log y)^{2}}{\log y-1}\).

Short Answer

Expert verified
\(\frac{d y}{d x}=\frac{(\log y)^{2}}{\log y-1}\)

Step by step solution

01

- Differentiation

Start by differentiating both sides of the equation \(e^{y}=y^{x}\) with respect to \(x\). On the left side, the derivative of \(e^{y}\) with respect to \(x\) is \(e^{y}(\frac{dy}{dx})\) by the chain rule. On the right side, the derivative of \(y^{x}\) with respect to \(x\) is \((y^{x})(\log(y) + \frac{x}{y}(\frac{dy}{dx}))\) by the general power rule. So, differentiation gives us the equation \(e^{y}(\frac{dy}{dx}) = (y^{x})(\log(y) + \frac{x}{y}(\frac{dy}{dx}))\).
02

- Rearranging

Rearrange the equation above to solve for \(\frac{dy}{dx}\), to obtain \(\frac{dy}{dx} = \frac{y^{x}\log(y)}{e^{y}-(\frac{x}{y})(y^{x})}\).
03

- Substituting \(e^{y}\)

Note that from the original equation, we have \(e^{y} = y^{x}\). Inserting that on the denominator of the right hand side results in, \(\frac{dy}{dx} = \frac{y^{x}\log(y)}{y^{x} - \frac{x}{y}(y^{x})}\).
04

- Simplifying and Isolating \(\frac{dy}{dx}\)

Further simplification gives us the final result, \(\frac{dy}{dx} = \frac{(\log y)^{2}}{\log y - 1}\).

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Most popular questions from this chapter

\text { If } x=a \sec \theta, y=b \tan \theta \text { prove that } \frac{d^{2} y}{d x^{2}}=-\frac{b^{4}}{a^{2} y^{3}} .

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If \(y=\sin ^{-1} x\), then prove that (i) \(\left(1-x^{2}\right) y_{2}-x y_{1}=0\) (ii) \(\left(1-x^{2}\right) y_{n+2}-(2 n+1) x y_{n+1}-n^{2} y_{n}=0\)
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