Question

If x^y = e^x–y then $\frac{dy}{dx}$ is equal to –

• A. (1 + log x)^–1
• B. (1 + logx)^–2
• C. (log x) (1 + logx)^–2
• D. None of these

Answer 1

Answer: (C)

(log x) (1 + logx)^–2

Answer 2

x^y = e^{x – y} ⇒ y logx = x – y .........(i)

$\frac{y}{x}$ + logx $\frac{dy}{dx}$ = 1 – $\frac{dy}{dx}$

⇒$\frac{dy}{dx}$ (1+logx) = 1 – $\frac{y}{x}$

⇒ $\frac{dy}{dx}$= $\left( 1–\frac{y}{x} \right)$.$\frac{1}{(1 + \log x)}$

from (i) put value of y

= $\left\lbrack 1–\frac{x}{(1 + \log x)} \times \frac{1}{x} \right\rbrack$ $\left( \frac{1}{1 + \log x} \right)$

= $\frac{1 + \log x–1}{(1 + \log x)^{2}}$ = (logx). (1 + logx)^–2