# Find the derivatives of f{{\left({x}\right)}}=\frac{3}{{2}}{x}^{4}+{2}{x}^{3}-{5}{x}^{2}+{9}

Find the derivatives of $$\displaystyle f{{\left({x}\right)}}=\frac{3}{{2}}{x}^{4}+{2}{x}^{3}-{5}{x}^{2}+{9}$$

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wornoutwomanC

Find the derivative of the following via implicit differentiation:
$$\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left( f{{\left({x}\right)}}\right)}=\frac{d}{{\left.{d}{x}\right.}}{\left({9}-{5}{x}^{2}+{2}{x}^{3}+\frac{{{3}{x}^{4}}}{{2}}\right)}$$
Using the chain rule, $$\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left( f{{\left({x}\right)}}\right)}=\frac{{{d} f{{\left({u}\right)}}}}{{{d}{u}}}\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$$, where $$\displaystyle{u}={x}{\quad\text{and}\quad}\frac{d}{{{d}{u}}}{\left( f{{\left({u}\right)}}\right)}={f}'{\left({u}\right)}$$:
$$\displaystyle{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}\right)}\right)}{f}'{\left({x}\right)}=\frac{d}{{\left.{d}{x}\right.}}{\left({9}-{5}{x}^{2}+{2}{x}^{3}+\frac{{{3}{x}^{4}}}{{2}}\right)}$$
The derivative of x is 1:
$$\displaystyle{1}{f}'{\left({x}\right)}=\frac{d}{{\left.{d}{x}\right.}}{\left({9}-{5}{x}^{2}+{2}{x}^{3}+\frac{{{3}{x}^{4}}}{{2}}\right)}$$
Differentiate the sum term by term and factor out constants:
$$\displaystyle{f}'{\left({x}\right)}=\frac{d}{{\left.{d}{x}\right.}}{\left({9}\right)}-{5}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{2}\right)}\right)}+{2}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{3}\right)}\right)}+\frac{3}{{2}}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{4}\right)}\right)}$$
The derivative of 9 is zero:
$$\displaystyle{f}'{\left({x}\right)}=-{5}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{2}\right)}\right)}+{2}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{3}\right)}\right)}+\frac{3}{{2}}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{4}\right)}\right)}+{0}$$
Simplify the expression:
$$\displaystyle{f}'{\left({x}\right)}=-{5}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{2}\right)}\right)}+{2}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{3}\right)}\right)}+\frac{3}{{2}}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{4}\right)}\right)}$$
Use the power rule, $$\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{n}\right)}={n}{x}^{{{n}-{1}}}$$, where n = 2.
$$\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{2}\right)}={2}{x}:$$
$$\displaystyle{f}'{\left({x}\right)}={2}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{3}\right)}\right)}+\frac{3}{{2}}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{4}\right)}\right)}-{5}{2}{x}$$
Simplify the expression:
$$\displaystyle{f}'{\left({x}\right)}=-{10}{x}+{2}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{3}\right)}\right)}+\frac{3}{{2}}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{4}\right)}\right)}$$
Use the power rule, $$\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{n}\right)}={n}{x}^{{{n}-{1}}}$$, where n = 3.
$$\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{3}\right)}={3}{x}^{2}:$$
$$\displaystyle{f}'{\left({x}\right)}=-{10}{x}+\frac{3}{{2}}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{4}\right)}\right)}+{2}{3}{x}^{2}$$
Simplify the expression:
$$\displaystyle{f}'{\left({x}\right)}=-{10}{x}+{6}{x}^{2}+\frac{3}{{2}}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{4}\right)}\right)}$$
Use the power rule, $$\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{n}\right)}={n}{x}^{{{n}-{1}}}$$, where n = 4.
$$\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{4}\right)}={4}{x}^{3}:$$
$$\displaystyle{f}'{\left({x}\right)}=-{10}{x}+{6}{x}^{2}+\frac{3}{{2}}{4}{x}^{3}$$
Simplify the expression:
$$\displaystyle{f}'{\left({x}\right)}=-{10}{x}+{6}{x}^{2}+{6}{x}^{3}$$
Simplify the expression:
$$\displaystyle={2}{x}{\left(-{5}+{3}{x}+{3}{x}^{2}\right)}$$