Use the forward$-$difference formula to approximate the derivative of $f(x)=\frac{x^3}{5}$ at $x_0=1$ using $h=0\mathrm{.}5$

a$)0\mathrm{.}75$

b$)0\mathrm{.}45$ c$)0\mathrm{.}60$

d$)0,95$

a$)$ None of these