Question

# Find the first and second derivatives. y=-x^{2}+3

Derivatives
Find the first and second derivatives.
$$\displaystyle{y}=-{x}^{{{2}}}+{3}$$

2021-05-05
Step 1
The given function is $$\displaystyle{y}=-{x}^{{{2}}}+{3}$$.
Formula:
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}^{{{n}}}\right)}={n}{x}^{{{n}-{1}}}$$
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({f}+{g}\right)}={\frac{{{d}{f}}}{{{\left.{d}{x}\right.}}}}+{\frac{{{d}{g}}}{{{\left.{d}{x}\right.}}}}$$
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({k}\right)}={0}$$ where k is constant.
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({k}{f}\right)}={k}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({f}\right)}$$
Step 2
Obtain the first derivative of y as follows.
$$\displaystyle{y}=-{x}^{{{2}}}+{3}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left(-{x}^{{{2}}}+{3}\right)}$$
$$\displaystyle={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left(-{x}^{{{2}}}\right)}+{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({3}\right)}$$
$$\displaystyle=-{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}^{{{2}}}\right)}+{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({3}\right)}$$
$$\displaystyle=-{2}{x}^{{{2}-{1}}}+{0}$$
$$\displaystyle=-{2}{x}^{{{1}}}$$
=-2x
Step 3
Obtain the second derivative as follows.
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=-{2}{x}$$
$$\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\right)}$$
$$\displaystyle={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left(-{2}{x}\right)}$$
$$\displaystyle=-{2}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}\right)}$$
$$\displaystyle=-{2}\times{1}{x}^{{{1}-{1}}}$$
$$\displaystyle=-{2}{x}^{{{0}}}$$
=-2