Question

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

Answer 1

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
\(\displaystyle{g}'{\left({2}\right)}=-{\frac{{{2}}}{{{\left({2}\right)}^{{{3}}}}}}\)
\(\displaystyle=-{\frac{{{2}}}{{{8}}}}\)
\(\displaystyle=-{\frac{{{1}}}{{{4}}}}\)
\(\displaystyle{g}'{\left(\sqrt{{{3}}}\right)}=-{\frac{{{2}}}{{{\left(\sqrt{{{3}}}\right)}^{{{3}}}}}}\)
\(\displaystyle=-{\frac{{{2}}}{{{3}\sqrt{{{3}}}}}}\)