Question

# f(x)=(8x-1)(x^{2}+4x+7)(x^{3}-5) Find the derivatives.

Derivatives
$$\displaystyle{f{{\left({x}\right)}}}={\left({8}{x}-{1}\right)}{\left({x}^{{{2}}}+{4}{x}+{7}\right)}{\left({x}^{{{3}}}-{5}\right)}$$
Find the derivatives.

2021-02-26

Given,
$$\displaystyle{f{{\left({x}\right)}}}={\left({8}{x}-{1}\right)}{\left({x}^{{{2}}}+{4}{x}+{7}\right)}{\left({x}^{{{3}}}-{5}\right)}$$
On simplification, we get
$$\displaystyle{f{{\left({x}\right)}}}={\left({8}{x}^{{{3}}}+{32}{x}^{{{2}}}+{56}{x}-{x}^{{{2}}}-{4}{x}-{7}\right)}{\left({x}^{{{3}}}-{5}\right)}$$
$$\displaystyle={8}{x}^{{{6}}}+{32}{x}^{{{5}}}+{56}{x}^{{{4}}}-{x}^{{{5}}}-{4}{x}^{{{4}}}-{7}{x}^{{{3}}}-{40}{x}^{{{3}}}-{160}{x}^{{{2}}}-{280}{x}+{5}{x}^{{{2}}}+{20}{x}+{35}$$
$$\displaystyle={8}{x}^{{{6}}}+{31}{x}^{{{5}}}+{52}{x}^{{{4}}}-{47}{x}^{{{3}}}-{155}{x}^{{{2}}}-{260}{x}+{35}$$
Now differentiating with respect to x, we get
$$\displaystyle{f}'{\left({x}\right)}={8}{\left({6}{x}^{{{5}}}\right)}+{31}{\left({5}{x}^{{{4}}}\right)}+{52}{\left({4}{x}^{{{3}}}\right)}-{47}{\left({3}{x}^{{{2}}}\right)}-{155}{\left({2}{x}\right)}-{260}{\left({1}\right)}+{0}$$
$$\displaystyle{\left[{\frac{{{d}{\left({x}^{{{n}}}\right)}}}{{{\left.{d}{x}\right.}}}}={n}{x}^{{{n}-{1}}}{\frac{{{d}{\left({c}\right)}}}{{{\left.{d}{x}\right.}}}}={0},{c}\in{R}\right]}$$
$$\displaystyle={48}{x}^{{{5}}}+{155}{x}^{{{4}}}+{208}{x}^{{{3}}}-{141}{x}^{{{2}}}-{310}{x}-{260}$$