Question

logarithmic differentiation of x^y + y^x​

Answer 1

Answer:

Step-by-step explanation:

y′=y(lnf(x))′=f(x)(lnf(x))′. The derivative of the logarithmic function is called the logarithmic derivative of the initial function y=f(x). y=u(x)v(x), where u(x) and v(x) are differentiable functions of x.

Answer 2

Answer:

${x}^{y} + {y}^{x} \\ taking \: log \\ ylog {x} + xlog {y} \\ \\ differentiating \: w \: r \: to \: x \: 1st \: term \\ \frac{dy}{dx} \: logx + y \: \frac{1}{x} \\ \\ \frac{dy}{dx} logx + \frac{y}{x} \\ \\ diff. \: second \: term \: \\ \: xlogy \\ 1log \: y \: + x \: \frac{1}{y} \\ logy + \frac{x}{y}$

$\frac{dy}{dx} logx + \frac{y}{x}+logy + \frac{x}{y}$