Question

# Find the first and second derivatives. y=\frac{7x^{5}}{5}-2x+9e^{x} \frac{dy}{dx}=

Derivatives
Find the first and second derivatives.
$$\displaystyle{y}={\frac{{{7}{x}^{{{5}}}}}{{{5}}}}-{2}{x}+{9}{e}^{{{x}}}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=$$

2021-06-09
Step 1
The given function is $$\displaystyle{y}={\frac{{{7}{x}^{{{5}}}}}{{{5}}}}-{2}{x}+{9}{e}^{{{x}}}$$
Obtain the first derivative as follows.
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\frac{{{7}{x}^{{{5}}}}}{{{5}}}}-{2}{x}+{9}{e}^{{{x}}}\right)}$$
$$\displaystyle={\frac{{{35}{x}^{{{4}}}}}{{{5}}}}-{2}+{9}{e}^{{{x}}}$$
$$\displaystyle={7}{x}^{{{4}}}-{2}+{9}{e}^{{{x}}}$$
Step 2
Obtain the second derivative as follows.
$$\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\right)}$$
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({7}{x}^{{{4}}}-{2}+{9}{e}^{{{x}}}\right)}$$
$$\displaystyle={28}{x}^{{{3}}}+{9}{e}^{{{x}}}$$
Thus, the first derivative is $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={7}{x}^{{{4}}}-{2}+{9}{e}^{{{x}}}$$
The second derivative is $$\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}={28}{x}^{{{3}}}+{9}{e}^{{{x}}}$$