Răspuns:
1,f `(x)=4x³/4-3x²/3+4*6x⁵/3=
x³-x²+24x⁵/3=
8x⁵+x³-x²
4.f `(x)=(5ˣ+3ˣ*2ˣ-3²*2²) `=
5ˣln5+3ˣln3*2ˣ+2ˣln2*3ˣ
5. f `(x)=[(3x²-5)(x²)-2x(x³-5x+1)]/x⁴=
(3x⁴-5x²-2x⁴+5x²-x)/x⁴=
(x⁴-x)/x⁴=(x³-1)/x³
7) f `(x)=7cosx-4x
8) f `(x)=x`lnx+x*ln `X-x `+3 `=
lnx+x/x-1+0=
lnx+1
Explicație pas cu pas:
