Using the definition, calculate the derivatives of the function. Then find the values of the derivatives as specified. g(t)=\frac{1}{t^{2}}. g'(-1), g'(2), g'(\sqrt{3})

nitraiddQ 2021-05-26 Answered
Using the definition, calculate the derivatives of the function. Then find the values of the derivatives as specified.
\(\displaystyle{g{{\left({t}\right)}}}={\frac{{{1}}}{{{t}^{{{2}}}}}}.{g}'{\left(-{1}\right)},{g}'{\left({2}\right)},{g}'{\left(\sqrt{{{3}}}\right)}\)

Expert Community at Your Service

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

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Macsen Nixon
Answered 2021-05-27 Author has 23531 answers
Step 1
The given function is,
\(\displaystyle{g{{\left({t}\right)}}}={\frac{{{1}}}{{{t}^{{{2}}}}}}\)
To find the derivative of the function, we use the power rule of differentiation,
\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}^{{{n}}}\right)}={n}{x}^{{{n}-{1}}}\)
Step 2
Applying differentiation on both sides of the function, we get
\(\displaystyle{g}'{\left({t}\right)}={\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({\frac{{{1}}}{{{t}^{{{2}}}}}}\right)}\)
\(\displaystyle={\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({t}^{{-{2}}}\right)}\)
\(\displaystyle=-{2}{t}^{{-{2}-{1}}}\)
\(\displaystyle=-{2}{t}^{{-{3}}}\)
\(\displaystyle=-{\frac{{{2}}}{{{t}^{{{3}}}}}}\)
Therefore, the derivative of the given function is \(\displaystyle{g}'{\left({t}\right)}=-{\frac{{{2}}}{{{t}^{{{3}}}}}}\)
Step 3
Finding the value of the derivative of the given function for the specified value of the variable, we get
\(\displaystyle{g}'{\left(-{1}\right)}=-{\frac{{{2}}}{{{\left(-{1}\right)}^{{{3}}}}}}\)
\(\displaystyle=-{\frac{{{2}}}{{-{1}}}}\)
=2
\(\displaystyle{g}'{\left({2}\right)}=-{\frac{{{2}}}{{{\left({2}\right)}^{{{3}}}}}}\)
\(\displaystyle=-{\frac{{{2}}}{{{8}}}}\)
\(\displaystyle=-{\frac{{{1}}}{{{4}}}}\)
\(\displaystyle{g}'{\left(\sqrt{{{3}}}\right)}=-{\frac{{{2}}}{{{\left(\sqrt{{{3}}}\right)}^{{{3}}}}}}\)
\(\displaystyle=-{\frac{{{2}}}{{{3}\sqrt{{{3}}}}}}\)
Not exactly what you’re looking for?
Ask My Question
39
 

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 2021-06-04
Using the definition, calculate the derivatives of the functions. Then find the values of the derivatives as specified.
\(\displaystyle{k}{\left({z}\right)}={\frac{{{1}-{z}}}{{{2}{z}}}}.{k}'{\left(-{1}\right)},{k}'{\left({1}\right)},{k}'{\left(\sqrt{{{2}}}\right)}\)
asked 2021-02-14
Calculate the derivatives of the functions. Then find the values of the derivatives as specified.
\(\displaystyle{f{{\left({x}\right)}}}={4}-{x}^{{{2}}};{f}'{\left(-{3}\right)},{f}'{\left({0}\right)},{f}'{\left({1}\right)}\)
asked 2021-09-28
Using the definition, calculate the derivatives of the function. Then find the values of the derivatives as specified.
\(\displaystyle{g{{\left({t}\right)}}}={\frac{{{1}}}{{{t}^{{{2}}}}}};{g}'{\left(-{1}\right)},{g}'{\left({2}\right)},{g}'{\left(\sqrt{{{3}}}\right)}\)
asked 2021-09-28
Using the definition, calculate the derivatives of the functions. Then find the values of the derivatives as specified.
\(\displaystyle{k}{\left({z}\right)}={\frac{{{1}-{z}}}{{{2}{z}}}};{k}'{\left(-{1}\right)},{k}'{\left({1}\right)},{k}'{\left(\sqrt{{{2}}}\right)}\)
asked 2021-11-21
Calculate all partial derivatives of the function
\(\displaystyle{g{{\left({x},{y},{z}\right)}}}={\frac{{{x}{\sin{{\left({y}\right)}}}}}{{{z}^{{{2}}}}}}\)
asked 2021-06-06
Find derivatives for the functions. Assume a, b, c, and k are constants.
\(\displaystyle{g{{\left({x}\right)}}}={\frac{{{x}^{{{2}}}+\sqrt{{{x}}}+{1}}}{{{x}^{{{\frac{{{3}}}{{{2}}}}}}}}}\)
asked 2021-04-18
Find the derivatives of the following functions.
\(\displaystyle{g{{\left({t}\right)}}}={{\sin{{h}}}^{{-{1}}}\sqrt{{{t}}}}\)
...