How do you compute the value of the right derivative of $f(x)= \sin (x)^{\cos (x)} +\cos (x)^{\sin (x)}$ when $x=0$. How do you compute the value of the right derivative of $f(x)= \sin (x)^{\cos (x)} +\cos (x)^{\sin (x)}$ when $x=0$. I'm trying to learn calculus so some explanations wouldn't be so bad. I got stuck computing the limit of $\sin (x)^{\cos (x)} \cdot \big( \frac{\cos ^2 (x)}{sin (x)} - \sin (x) \cdot \ln (\sin (x)\big)$ as $x \rightarrow 0$. Sorry for the grammar mistakes but I'm not English.
 A: $$\lim_{x\to 0}\frac{(\sin x)^{\cos x}-0}x=\lim_{x\to 0}\frac{(\sin x)^{1+o(x)}-0}x=1$$
and
$$\lim_{x\to 0}\frac{(\cos x)^{\sin x}-1}x=\lim_{x\to 0}\frac{(1-\frac{x^2}2+o(x^2))^{\sin x}-1}x
\\=\lim_{x\to 0}\frac{(1-\sin x\frac{x^2}2+o(\sin x\,x^2))-1}x
\\=0.$$
Note that we exclude $x<0$ from the domain, so that the limits need not be right-hand.
A: You should learn logarithmic differentiation:
$$h(x)={(\sin{x})}^{\cos{x}}$$
$$\ln{(h(x))}=\cos{x} \cdot \ln{\sin{x}}$$
$$\frac{h'(x)}{h(x)}=-\sin{x} \cdot \ln{\sin{x}}+\frac{\cos^2{x}}{\sin{x}}$$
$$h'(x)={(\sin{x})}^{\cos{x}} \left(-\sin{x} \cdot \ln{\sin{x}}+\frac{\cos^2{x}}{\sin{x}}\right)$$
Do this for $g(x)={(\cos{x})}^{\sin{x}}$:
$$g'(x)= {(\cos{x})}^{\sin{x}} \left( \cos{x} \cdot \ln{\cos{x}} -\frac{ \sin^2{x}}{\cos{x}} \right)$$
To sum it up,
$$f(x)=g(x)+h(x) \implies f'(x)=g'(x)+h'(x)$$
$$f'(x)={(\sin{x})}^{\cos{x}} \left(-\sin{x} \cdot \ln{\sin{x}}+\frac{\cos^2{x}}{\sin{x}}\right)+{(\cos{x})}^{\sin{x}} \left( \cos{x} \cdot \ln{\cos{x}} -\frac{ \sin^2{x}}{\cos{x}} \right)$$
Using the Maclaurin series approximations of $\sin{x}\approx x$ and $\cos{x} \approx 1-\frac{x^2}{2}$ near $x=0$:
$$\lim_{x \to 0^+} f'(x)=x^1 \left(0+ \frac{1-x^2+\frac{x^4}{4}}{x}\right)+1\left(0-0\right)=\boxed{1}$$
