Răspuns:
f)f(x)=2ˣ+5x
f `(x)=2ˣln2+5ˣln5
g)f(x)=lnx-x+1
f `(x)=(lnx) `-x`+1`=
1/x-1+0=1/x-1
h)f(x)=2sinx+3cosx
f `(x)=(2sinx) `+(3cosx) `=
2cosx-3sinx
i f(x)=x²+2⁻⁻ˣ
f `(x)=(x²) `+(2⁻⁻ˣ) `=
2x-2⁻⁻ˣln2
j)f(x)=tgx+ctgx=.>
f `(x)=(tgx) `+(ctg x) `=
1/cos²x-1/sin²x=
(sin²x-cos²x)/sin²x*cos²x=
-cos2xsin²c*cos²x
Explicație pas cu pas:
