Răspuns:
f: IR -> IR , f(x ) = 2^(- 2x+1)
f'(x) = d/dx [2^-(2x+1)]
Folosim:
d/dx=[f(g)] = d/dg[f(g)]× d/dg (g)
g= - 2x+1
f'(x) = d/dg (2^g) × d/dx (-2x+1)
Folosim :
d/dx(a^x)= ln(a) × a^x
d/dg(2^g)= ln(2)×2^g
Folosim :
d/dx(f+g) =d/dx(f) +d/dx(g)
d/dx(-2x+1)=d/dx(-2x)+ d/dx(1)
d/dx(-2x+1)= -2+0
d/dx(-2x+1)= -2
f'(x) = d/dg (2^g) × d/dx (-2x+1)
f'(x) = ln(2)×2^g × (-2)
g= - 2x+1
f'(x)= ln(2)×2^(-2x+1)×(-2)
f'(x) = - ln(2)×2^(2n+2)
Explicație pas cu pas: