Вычислить производные функций! (с решением )

Решите задачу:

$y'(x^9-3x^5- \frac{3}{x^4}+2) =(x^9-3x^5-3x^{-4}+2)'=9x^8-3x^4+12x^{-5}=\\\\ 9x^8-3x^4+ \frac{12}{x^5}= \frac{9x^{13}-15x^9+12}{x^5}\\\\ ***********************************\\\\ y'[(x+1) \sqrt{x}]=(x+1)' \sqrt{x}+(x+1)( \sqrt{x})'=1\cdot \sqrt{x}+(x+1)\cdot \frac{1}{2 \sqrt{x}}=\\\\ = \sqrt{x}+ \frac{x+1}{2 \sqrt{x}}= \frac{2x+x+1}{2 \sqrt{x}}= \frac{3x+1}{2 \sqrt{x}}\\\\ ***********************************$

$y'(tg\ x\cdot \sin(2x+5))=(tg\ x)'(\sin(2x+5))+tg\ x\cdot(\sin(2x+5))'= \\\\\ \frac{\sin(2x+5)}{\cos^2x}+2tg\ x \cdot \cos(2x+5)\\\\ ******************************************\\\\ y'(15x^4-4x^5)^{101})=101(15x^4-4x^5)^{100}\cdot(60x^3-20x^4)$

$********************************************$

$y'( \sqrt{3x^2-6x})=((3x^2-6x)^{ \frac{1}{2}})'= \frac{1}{2}(3x^2-6x)^{- \frac{1}{2}}(6x-6)=\\\\ \frac{1}{2}\cdot 6(x-1)(3x^2-6x)^{- \frac{1}{2}}= \frac{3(x-1)}{ \sqrt{3x^2-6x}}= \frac{3(x-1)}{ \sqrt{3} \sqrt{x(x-2)}}= \frac{ \sqrt{3}(x-1)}{ \sqrt{x(x-2)}}$