91Ó°ÊÓ

The given functions are following.

$$\begin{gathered} f\left(x\right)={\mathrm{Ï€}}^2 \\ f\left(x\right)={\left(\sqrt[11]{x}\right)}^{-2} \\ f\left(x\right)=\frac{1}{5x} \\ f\left(x\right)=se{c}^2(3x+1) \end{gathered}$$

Derivative of $f\left(x\right)={\mathrm{Ï€}}^2.$

$$\begin{gathered} \frac{d}{dx}f\left(x\right)=\frac{d}{dx}{\mathrm{Ï€}}^2 \\ \frac{d}{dx}f\left(x\right)=0 \end{gathered}$$

antiderivative of $f\left(x\right)={\mathrm{Ï€}}^2$

$$\begin{gathered} ∫f\left(x\right)dx=∫{\mathrm{π}}^{2}dx \\ ∫f\left(x\right)dx={\mathrm{π}}^{2}x+C \end{gathered}$$

Derivative of$f\left(x\right)={\left(\sqrt[11]{x}\right)}^{-2}.$

$$\begin{gathered} \frac{d}{dx}f\left(x\right)=\frac{d}{dx}{\left(\sqrt[11]{x}\right)}^{-2} \\ \frac{d}{dx}f\left(x\right)=\frac{-2}{{\left(\sqrt[11]{x}\right)}^3}\frac{d}{dx}\left(\sqrt[11]{x}\right) \\ \frac{d}{dx}f\left(x\right)=\frac{-2}{{\left(\sqrt[11]{x}\right)}^3}×\frac{1}{11x^{{\displaystyle \frac{10}{11}}}} \\ \frac{d}{dx}f\left(x\right)=\frac{-2}{11x^{{\displaystyle \frac{15}{11}}}} \end{gathered}$$

antiderivative of$f\left(x\right)={\left(\sqrt[11]{x}\right)}^{-2}.$

$$\begin{gathered} ∫f\left(x\right)dx=∫{\left(\sqrt[11]{x}\right)}^{-2}dx \\ ∫f\left(x\right)dx=∫x^{\frac{-2}{11}}dx \\ ∫f\left(x\right)dx=\frac{11}{9}{x}^{\frac{9}{11}}+C \end{gathered}$$

Derivative of$f\left(x\right)=\frac{1}{5x}$

$$\begin{gathered} \frac{d}{dx}f\left(x\right)=\frac{d}{dx}\frac{1}{5x} \\ \frac{d}{dx}f\left(x\right)=\frac{1}{5}\frac{d}{dx}{x}^{-1} \\ \frac{d}{dx}f\left(x\right)=\frac{1}{5}\left(-1x^{-2}\right) \\ \frac{d}{dx}f\left(x\right)=\frac{-1}{5x^2} \end{gathered}$$

antiderivative of $f\left(x\right)=\frac{1}{5x}.$

localid="1648626492091" $$\begin{gathered} ∫f\left(x\right)dx=∫\frac{1}{5x}dx \\ ∫f\left(x\right)dx=\frac{1}{5}∫\frac{1}{x}dx \\ ∫f\left(x\right)dx=\frac{1}{5}\log x+C \end{gathered}$$

Derivative of $f\left(x\right)=se{c}^2(3x+1)$

localid="1648626533667" $$\begin{gathered} \frac{d}{dx}f\left(x\right)=\frac{d}{dx}se{c}^2(3x+1) \\ =2sec(3x+1)\frac{d}{dx}sec(3x+1) \\ =2sec(3x+1)·sec(3x+1)·\tan (3x+1)·3 \\ =6se{c}^2(3x+1)\tan (3x+1) \end{gathered}$$

antiderivative of $f\left(x\right)=se{c}^2(3x+1).$

$$\begin{gathered} ∫f\left(x\right)dx=∫se{c}^2(3x+1)dx \\ ∫f\left(x\right)dx=\frac{1}{3}\tan (3x+1)+C \end{gathered}$$