$$2^{1}+b^{2}+-1$$
$b^{2}+1$
$$2+b^{2}-1$$
$$1+b^{2}$$
Show Solution
Hide Solution
$2b$
$$\frac{\mathrm{d}}{\mathrm{d}b}(2+b^{2}-1)$$
$$\frac{\mathrm{d}}{\mathrm{d}b}(1+b^{2})$$
$$2b^{2-1}$$
$$2b^{1}$$
$$2b$$