$\displaystyle \left[\dfrac{x+\sin x}{x-\sin x}\right]'$

$\displaystyle \left[\dfrac{x+\sin x}{x-\sin x}\right]'= \dfrac{[x+\sin x]'(x-\sin x)-(x+\sin x)[x-\sin x]'}{(x-\sin x)^2}= \dfrac{(1+\cos x)(x-\sin x)-(x+\sin x)(1-\cos x)}{(x-\sin x)^2}=\\ \qquad\displaystyle \dfrac{(x-\sin x+x\cos x-\sin x\cos x)-(x+\sin x-x\cos x-\sin x\cos x)}{(x-\sin x)^2}= \bbox[#FFECB3,5px]{2\dfrac{x\cos x-\sin x}{(x-\sin x)^2}}$

$\displaystyle \left[\dfrac{e^x+e^{-x}}{2}\right]'$

$\displaystyle \left[\dfrac{e^x+e^{-x}}{2}\right]'= \bbox[#FFECB3,5px]{\dfrac{e^x-e^{-x}}{2}}$

Tip: these are the hyperbolic cosine and sine

$$\sinh x=\dfrac{e^x-e^{-x}}{2}$$ $$\cosh x=\dfrac{e^x+e^{-x}}{2}$$

and satisfy

$$[\sinh x]'=\cosh x$$ $$[\cosh x]'=\sinh x$$

$\displaystyle \left[\dfrac{e^x-e^{-x}}{e^x+e^{-x}}\right]'$

$\displaystyle \left[\dfrac{e^x-e^{-x}}{e^x+e^{-x}}\right]'= \dfrac{[e^x-e^{-x}]'(e^x+e^{-x})-(e^x-e^{-x})[e^x+e^{-x}]'}{(e^x+e^{-x})^2}= \dfrac{(e^x+e^{-x})(e^x+e^{-x})-(e^x-e^{-x})(e^x-e^{-x})}{(e^x+e^{-x})^2}=\\ \qquad\displaystyle \dfrac{(e^{2x}+2+e^{-2x})-(e^{2x}-2+e^{-2x})}{(e^x+e^{-x})^2}= \bbox[#FFECB3,5px]{\dfrac{4}{(e^x+e^{-x})^2}}$

Tip: this is the hyperbolic tangent

$$\tanh x=\dfrac{\sinh x}{\cosh x}=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$$ and its derivative is

$$[\tanh x]'=\dfrac{4}{(e^x+e^{-x})^2}=\dfrac{1}{\cosh^2(x)}$$

Quite similar to the usual trigonometric functions, aren't they?

$\displaystyle \left[\sqrt{3x\tan(x^3)}\right]'$

$\displaystyle \left[\sqrt{3x\tan(x^3)}\right]'= \dfrac{1}{2\sqrt{3x\tan(x^3)}}[3x\tan(x^3)]'= \dfrac{3}{2\sqrt{3x\tan(x^3)}}[x\tan(x^3)]'= \dfrac{\sqrt{3}}{2\sqrt{x\tan(x^3)}}(\tan(x^3)+x(1+\tan^2(x^3))3x^2)=\\ \qquad\displaystyle \bbox[#FFECB3,5px]{\dfrac{\sqrt{3}(3x^3(1+\tan^2(x^3))+\tan(x^3))}{2\sqrt{x\tan(x^3)}}}$

$\displaystyle \left[e^{\sin(4x)}\right]'$

$\displaystyle \left[e^{\sin(4x)}\right]'= e^{\sin(4x)}[\sin(4x)]'= \bbox[#FFECB3,5px]{4e^{\sin(4x)}\cos(4x)}$