Question

# Find the derivatives of all orders of the functions y=(4x^{2}+3)(2-x)x

Derivatives
Find the derivatives of all orders of the functions $$\displaystyle{y}={\left({4}{x}^{{{2}}}+{3}\right)}{\left({2}-{x}\right)}{x}$$

2021-03-26
Step 1
We first expand the expression
$$\displaystyle{y}={\left({4}{x}^{{{2}}}+{3}\right)}{\left({2}-{x}\right)}{x}$$
$$\displaystyle{y}={\left({8}{x}^{{{2}}}-{4}{x}^{{{3}}}+{6}-{3}{x}\right)}{x}$$
$$\displaystyle{y}={8}{x}^{{{3}}}-{4}{x}^{{{4}}}+{6}{x}-{3}{x}^{{{2}}}$$
$$\displaystyle{y}=-{4}{x}^{{{4}}}+{8}{x}^{{{3}}}-{3}{x}^{{{2}}}+{6}{x}$$
Step 2
Then we find the derivatives using the power rule
$$\displaystyle{y}=-{4}{x}^{{{4}}}+{8}{x}^{{{3}}}-{3}{x}^{{{2}}}+{6}{x}$$
$$\displaystyle{y}^{{{\left({1}\right)}}}=-{16}{x}^{{{3}}}+{24}{x}^{{{2}}}-{6}{x}+{6}$$
$$\displaystyle{y}^{{{\left({2}\right)}}}=-{48}{x}^{{{2}}}+{48}{x}-{6}$$
$$\displaystyle{y}^{{{\left({3}\right)}}}=-{96}{x}+{48}$$
$$\displaystyle{y}^{{{\left({4}\right)}}}=-{96}$$
$$\displaystyle{y}^{{{\left({5}\right)}}}={0}$$
All other higher-order derivatives will be 0