Question

Differentiate f(x)=tan^-1(1-x/1+x)
Differentiate x^x

Answer 1

i)

$\frac{d(arctan\frac{1-x}{1+x})}{dx}=$ $\frac{1}{(\frac{1-x}{1+x})^{2}+1}\frac{d}{dx}\frac{1-x}{1+x}$

=$\frac{1}{(\frac{1-x}{1+x})^{2}+1}\frac{(-1)(1+x)-(1)(1-x)}{(1+x)^{2}}$

=$\frac{1}{(\frac{1-x}{1+x})^{2}+1}\frac{-2}{(1+x)^{2}}$

simplifying we get

d(tan^-1(1-x/1+x) = $\frac{-1}{1+x^{2}}$

ii) d(x^x)/dx = $\frac{d}{dx}e^{ln(x) x}$

= e^xln(x) $\frac{d}{dx}xlnx$

=$e^{xlnx}(1*lnx+x*\frac{1}{x})$

= $e^{xlnx}(lnx+1)$

$x^{x}(ln x+1)$