What would be the derivative of this? Could anyone help me find the derivative of:- 
$$ (x^{e^{\arctan(x)}})(x^{\ln x} + (\ln x)^{(x^{\sin x})}) $$
Showing some steps would be really useful.
 A: $$\ f(x)=(x^{e^{\arctan(x)}})(x^{\ln x}+(\ln x)^{(x^{\sin x})})$$
Taking the log on both sides as suggested in the comments, you get:
$$\ \ln[f(x)]=e^{\arctan (x)}\cdot\ln x+\ln[x^{\ln x}+(\ln x)^{x^{\sin x}}]$$
Now, before proceeding, let us compute the following derivatives:
$$\ g(x)=x^{\ln x}=e^{\ln^2 x}\implies g'(x)=\frac{2\ln x}{x}e^{\ln^2 x}=\frac{2\ln x}{x}x^{\ln x}=2x^{\ln (x)-1}\cdot \ln x$$
$$\ h(x)=x^{\sin x}=e^{\sin (x)\cdot \ln x}\implies h'(x)=[\cos(x)\ln x+\frac{\sin(x)}{x}]e^{\sin(x)\cdot \ln x}=$$
$$\ =[\cos(x)\ln x+\frac{\sin(x)}{x}]x^{\sin x}$$
$$\ k(x)=(\ln x)^{x^{\sin x}}=(\ln x)^{h(x)}=e^{h(x)\ln(\ln x)}$$
$$\implies k'(x)=[h'(x)\ln(\ln x)+\frac{h(x)}{x\ln x}]e^{h(x)\ln(\ln x)}=$$
$$\ =[h'(x)\ln(\ln x)+\frac{h(x)}{x\ln x}]k(x)$$
So now we can use these results to write $\ f'(x)$:
$$\ \frac{f'(x)}{f(x)}=\frac{e^{\arctan(x)}\ln x}{1+x^2}+\frac{e^{\arctan(x)}}{x}+\frac{g'(x)+k'(x)}{g(x)+k(x)}$$
A: $$(x^{e^{\arctan(x)}})(x^{\ln x} + (\ln x)^{(x^{\sin x})})\\=\exp(\exp(\arctan(x))\cdot\ln(x))\cdot\left(\exp(\ln^2(x))+\exp(\exp(\sin(x)\cdot\ln(x))\cdot\ln(\ln(x)))\right)$$
Now have fun using the product, addition, and chain rules of the differentiation; and that $(\exp(x))'=\exp(x)$, $(\ln(x))'=\frac{1}{x}$, $(\sin(x))'=\cos(x)$, and $(x^2)'=2x$. In this day and age, just make sure you know how to use those rules but let a computer do it for you.
