Question

Vă rog mult!!!!!!!!!!!!!!!

Answer 1

Răspuns:

La ambele exercitii se aplica derivata functiei compuse:

(f(g(x))'=f'(g(x))·g'(x)

Explicație pas cu pas:

Ex1

$a) ~f'(x)=(e^{2x+1})'=e^{2x+1}*(2x+1)'=2e^{2x+1}\\b)~f'(x)=((2x+1)^{7})'=7*(2x+1)^{6}*(2x+1)'=14(2x+1)^{6}.\\c)~f'(x)=((3-x)^{9})'=9(3-x)^{8}*(3-x)'=-9(3-x)^{8}\\d)~f'(x)=((x^{2}-3x+4)^{10})'=10(x^{2}-3x+4)^{9}*((x^{2}-3x+4)'=10(2x-3)*(x^{2}-3x+4)^{9}\\e)~f'(x)=(\sqrt{x^{2}+2x+7})'=\dfrac{1}{2\sqrt{x^{2}+2x+7}}*(x^{2}+2x+7)'= \dfrac{2x+2}{2\sqrt{x^{2}+2x+7}}=\dfrac{x+1}{\sqrt{x^{2}+2x+7}}\\f)~f'(x)=(ln(x^{2}+x+1))'=\dfrac{1}{x^{2}+x+1}*(x^{2}+x+1)'= \dfrac{2x+1}{x^{2}+x+1}$

$h)~f'(x)=(sin^{2}x)'=2sinx*(sinx)'=2sinxcosx=sin2x\\i) ~f'(x)=(sin3x)'=cos3x*(3x)'=3cos3x.$