Question

derivatele functiilor

f: D->R, f(x)=$2\sqrt{x} \cdot sinx$



f: D->R, f(x)=$(6x^{2} -5x)^{4}$

Answer 1

Salut,

$(2\sqrt x\cdot sinx)^{'}=2\cdot(\sqrt x\cdot sinx)^{'}=2\cdot (\sqrt x)^{'}\cdot sinx+2\cdot\sqrt x\cdot (sinx)^{'}=\\\\=2\cdot\dfrac{1}{2\cdot\sqrt x}\cdot sinx+2\cdot\sqrt x\cdot cosx=\dfrac{1}{\sqrt x}\cdot sinx+2\cdot\sqrt x\cdot cosx.$

[(6x² -- 5x)⁴ ] ' = 4·(6x² -- 5x)³·(6x² -- 5x) ' = 4·(6x² -- 5x)³·(2·6x -- 5) =

= 4·(6x² -- 5x)³·(12·x -- 5).

Green eyes.