Question

me pueden ayudar a resolver
son derivadas
producto.
f' =f'g + f'g'

cociente.
f= f' g - f . g'
__________
g²


#1........f(x)=(2x³+4x²+5x)(2x²+3x-2)

#2.....f(x)=(4x²-7×+7) (ײ+3×-4)

#3.....f(x)=3׳-8ײ+x-16
_______________
7x+5

#4......f(x)=12x²+9x-8
________________
x²+2x²-5x+1..

muchas gracias ...
solo gente seria...​

Answer 1

Respuesta:

$f(x)=(2x^3+4x^2+5x)(2x^2+3x-2)\\f'(x)=(6x^2+8x+5)(4x+3)\\f'(x)=24x^3+18x^2+32x^2+24x+20x+15\\f'(x)=24x^3+50x^2+44x+15$

$f(x)=(4x^2-7x+7)(x^2+3x-4)\\f'(x)=(8x-7)(2x+3)\\f'(x)=16x^2+24x-14x-21\\f'(x)=16x^2+10x-21$

$f(x)=\frac{3x^3-8x^2+x-16}{7x+5} \\\\f'(x)=\frac{(9x^2-16x+1)(7x+5)-(3x^3-8x^2+x-16)(7)}{(7x+5)^2} \\\\f'(x)=\frac{42x^3-11x^2-80x+117}{(7x+5)^2}$

$f(x)=\frac{12x^2+9x-8}{x^3+2x^2-5x+1} \\\\f'(x)=\frac{(24x+9)(x^3+2x^2-5x+1)-(12x^2+9x-8)(3x^2+4x-5)}{(x^3+2x^2-5x+1)^2} \\\\f'(x)=\frac{-12x^4-18x^3-54x^2+56x-31}{(x^3+2x^2-5x+1)^2}$