Question

помогите срочно !!!!!!!!!​

Answer 1

$y=24\ \ ,\ \ y'=0\\\\y=x\ \ ,\ \ y'=1\\\\y=25x\ \ ,\ \ y'=25\\\\y=x^3\ \ ,\ \ y'=3x^2\\\\y=7x^2+6\ \ ,\ \ y'=14x\\\\\\y=\dfrac{4}{6}\, x^6+\dfrac{4}{5}\, x^5+4x^4+4\ \ ,\ \ y'=4x^5+5x^4+16x^3\\\\\\y=(2x^4-3)(3x^2-4)\ \ ,\ \ y'=8x^3(3x^2-4)+6x(2x^4-3)=36x^5-32x^3-18x\\\\\\y=\dfrac{5x^3}{x+3}\ \ ,\ \ y'=\dfrac{15x^2(x+3)-5x^3}{(x+3)^2}=\dfrac{10x^3+45x^2}{(x+3)^2}=\dfrac{5x^2\, (2x+9)}{(x+3)^2}$