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рдкреНрд░рд╢реНрди
If \[y=x^x+x^{\frac{1}{x}}\] then \[\frac{\mathrm{d}y}{\mathrm{d}x}\] is equal to
рдкрд░реНрдпрд╛рдп
\[x^x\left(1+\log x\right)+x^{\frac{1}{x}}\frac{1}{x^2}\left(1-\log x\right)\]
\[\left(x^x+x^{\frac{1}{x}}\right)\left[1+\log x+\frac{1}{x^2}\left(1-\log x\right)\right]\]
\[\left(x^x+x^{\frac{1}{x}}\right)\left[\left(1+\log x\right)-\frac{1}{x^2}\left(1-\log x\right)\right]\]
\[x^{x}\left(1+\log x\right)-x^{\frac{1}{x}}\frac{1}{x^{2}}\left(1-\log x\right)\]
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рдЙрддреНрддрд░
\[x^x\left(1+\log x\right)+x^{\frac{1}{x}}\frac{1}{x^2}\left(1-\log x\right)\]
Explanation:
Let yтВБ = x╦г
∴ log yтВБ = x log x
Differentiating w.r.t. x, we get
\[\frac{1}{y_1}\frac{\mathrm{d}y_1}{\mathrm{d}x}=1+\log x\]
\[\begin{array} {cc}\therefore & \quad\frac{\mathrm{d}y_1}{\mathrm{d}x}=x^x\left(1+\log x\right) \end{array}\]
Let y2 = \[\frac{1}{X^X}\]
∴\[\begin{array} {cc} & \log y_2=\frac{1}{x}\log x \end{array}\]
Differentiating w.r.t.x, we get
\[\frac{1}{y_2}\frac{\mathrm{d}y_2}{\mathrm{d}x}=\frac{1}{x^2}-\frac{\log x}{x^2}\]
\[\begin{array} {ccc} & & & \\ & \therefore & & \frac{\mathrm{d}y_2}{\mathrm{d}x}=x^{\frac{1}{x}}\frac{1}{x^2}\left(1-\log x\right) \end{array}\]
\[y=x^x+x^ \frac1x\]
\[\therefore\quad\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y_{1}}{\mathrm{d}x}+\frac{\mathrm{d}y_{2}}{\mathrm{d}x}=x^{x}\left(1+\log x\right)+x^{\frac{1}{x}}\frac{1}{x^{2}}\left(1-\log x\right)\]
