Question

Derivati urmatoarele functii. Multumesc!!

Answer 1

Răspuns:

a) f(x) = e^(1+5x)

f'(x) = d/dx × e^(1+5x)

Folosim regula derivarii :

d/dx(f(g)) =d/d(g) (f(g)) ×d/dx(g)

f'(x) =d/dg (e^g) ×d/dx (1+5x)

Folosim formulele :

d/dx(e^x) =e^x

d/dx(f+g) = d/dx(f) + d/dx(g)

f'(x) = e^g ×[ d/dx(1) + d/dx(5x)]

f'(x) = e^g×(0+5)

f'(x) = e^g×5

inlocuieste g =1+5x

f'(x) = e^(1+5x)×5

f'(x) = 5e^(1+5x)

b) f(x) =sin^2x / 2+cos^2x

f'(x)= d/dx (sin^2x / 2+cos^2x)

f'(x) =[ d/dx (sin^2x) (2+cos^2x) - sin^2x×d/dx (2+cos^2x)] / (2+cos^2x)^2

f'(x) =[ 2sinxcosx ×(2+cos^2x) - sin^2 x (2cosx) (-sin x)] / (2+cos^2x)^2

f'(x) = [(sin 2x) (2+cos^2x)-(sin^2x) (2cos x)(-sin x)] / ((2+cos^x)^2)

f'(x) =[( sin 2x) (2+cos^2x)+(sin^2x) (sin 2x)] / ((2+cos^x)^2)

f'(x) =[( 2 sin 2x)+ (sin 2x)(cos^2x)+(sin^2x)(sin 2x)]/ ((2+cos^x)^2)

f'(x) ={( 2 sin 2x) +(sin 2x)[(cos^2x)+(sin^2x)]} / ((2+cos^x)^2)

f'(x) =[(2 sin 2x) +(sin 2x)×1] /

((2+cos^x)^2)

f'(x) = (3 sin 2x) / ((2+cos^x)^2)

c)

f(x) =rad( 1/ x-3)

f'(x) = d/dx rad( 1/ x-3)

f'(x) = d/dx 1/ rad(x-3)

f'(x) = - [d/dx rad (x-3)] / rad(x-3)^2

f'(x) = - [d/dg (rad g) ×d/dg (x-3)] /

rad(x-3)^2

Observatie :

d/dg (x-3)= d/dx(x) - d/dx(3)= 1-0= 1

f'(x) = - [(1/ 2rad g) ×1] / rad(x-3)^2

Inlocuim :g=x-3

f'(x)= - [(1/ 2rad (x-3)] / rad(x-3)^2

f'(x) = - 1 / 2 rad(x-3) × (x-3)

Explicație pas cu pas: