Question

# 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})

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

2021-05-27
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
$$\displaystyle{g}'{\left({2}\right)}=-{\frac{{{2}}}{{{\left({2}\right)}^{{{3}}}}}}$$
$$\displaystyle=-{\frac{{{2}}}{{{8}}}}$$
$$\displaystyle=-{\frac{{{1}}}{{{4}}}}$$
$$\displaystyle{g}'{\left(\sqrt{{{3}}}\right)}=-{\frac{{{2}}}{{{\left(\sqrt{{{3}}}\right)}^{{{3}}}}}}$$
$$\displaystyle=-{\frac{{{2}}}{{{3}\sqrt{{{3}}}}}}$$