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

amanf 2021-09-15 Answered

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

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

wornoutwomanC
Answered 2021-09-16 Author has 8112 answers

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:
Answer:
\(\displaystyle={2}{x}{\left(-{5}+{3}{x}+{3}{x}^{2}\right)}\)

Have a similar question?
Ask An Expert
33
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-09-02

how to find the derivative of \(\displaystyle f{{\left({u}\right)}}={9}{e}^{u}+{20}\)

asked 2021-09-13

Find the derivatives of \(\displaystyle{h}{\left({x}\right)}={\left(-{7}{\left({x}^{2}-{8}\right)}{2}\right)}{2}{x}\)

asked 2021-09-10

Define \(\displaystyle g{{\left({x},{y}\right)}}={x}^{2}+{y}^{2}-{4}{x}{y}+{3}{y}+{2}\)

asked 2021-05-19
For digits before decimals point, multiply each digit with the positive powers of ten where power is equal to the position of digit counted from left to right starting from 0.
For digits after decimals point, multiply each digit with the negative powers of ten where power is equal to the position of digit counted from right to left starting from 1.
1) \(10^{0}=1\)
2) \(10^{1}=10\)
3) \(10^{2}=100\)
4) \(10^{3}=1000\)
5) \(10^{4}=10000\)
And so on...
6) \(10^{-1}=0.1\)
7) \(10^{-2}=0.01\)
8) \(10^{-3}=0.001\)
9) \(10^{-4}=0.0001\)
asked 2021-06-13
Find derivatives for the functions. Assume a, b, c, and k are constants.
\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{{3}}}}}{{{9}}}}{\left({3}{\ln{{x}}}-{1}\right)}\)
asked 2021-05-09
Find all the second partial derivatives.
\(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{{4}}}{y}-{2}{x}^{{{5}}}{y}^{{{2}}}\)
\(\displaystyle{{f}_{{\times}}{\left({x},{y}\right)}}=\)
\(\displaystyle{{f}_{{{x}{y}}}{\left({x},{y}\right)}}=\)
\(\displaystyle{{f}_{{{y}{x}}}{\left({x},{y}\right)}}=\)
\(\displaystyle{{f}_{{{y}{y}}}{\left({x},{y}\right)}}=\)
asked 2021-05-04
Use the theorems on derivatives to find the derivatives of the following function:
\(\displaystyle{f{{\left({x}\right)}}}={3}{x}^{{{5}}}-{2}{x}^{{{4}}}-{5}{x}+{7}+{4}{x}^{{-{2}}}\)
...