91Ó°ÊÓ

The given function is$f\left(x\right)={\left(2x-1\right)}^3{\left(3x+1\right)}^2.$

To find the derivative we will use the product rule of differentiation.

So,

$$\begin{gathered} f\left(x\right)={\left(2x-1\right)}^3{\left(3x+1\right)}^2 \\ \begin{array}{rl}f\text{'}& \left(x\right)=(3x+1{)}^2\frac{d}{dx}(2x−1{)}^3+(2x−1{)}^3\frac{d}{dx}(3x+1{)}^2\\ f\text{'}& \left(x\right)=(3x+1{)}^{2}3(2x−1{)}^2\frac{d}{dx}(2x−1)+(2x−1{)}^{3}2(3x+1\left)\frac{d}{dx}\right(3x+1)\\ f\text{'}& \left(x\right)=6(3x+1{)}^2(2x−1{)}^2+6(2x−1{)}^3(3x+1)\\ f\text{'}& \left(x\right)=6(3x+1)(2x−1{)}^2(3x+1+2x−1)\\ f\text{'}& \left(x\right)=30x(3x+1)(2x−1{)}^2\end{array} \end{gathered}$$

To find the derivative we will use the quotient rule of differentiation.

So,

$$\begin{gathered} f\left(x\right)=\frac{{\left(2x-1\right)}^3}{{\left(3x+1\right)}^2} \\ \begin{array}{rl}f\text{'}& \left(x\right)=\frac{(3x+1{)}^2\frac{d}{dx}(2x−1{)}^3−(2x−1{)}^3\frac{d}{dx}(3x+1{)}^2}{{\left(\right(3x+1{)}^2)}^2}\\ f\text{'}& \left(x\right)=\frac{(3x+1{)}^2(2x−1{)}^2\frac{d}{dx}(2x−1)−(2x−1{)}^3(3x+1\left)\frac{d}{dx}\right(3x+1)}{{\left(\right(3x+1{)}^2)}^2}\\ f\text{'}& \left(x\right)=\frac{6(3x+1{)}^2(2x−1{)}^2−6(2x−1{)}^3(3x+1)}{{\left(\right(3x+1{)}^2)}^2}\\ f\text{'}& \left(x\right)=\frac{6(3x+1)(2x−1{)}^2(3x+1−(2x−1))}{(3x+1{)}^4}\\ f\text{'}& \left(x\right)=\frac{6(2x−1{)}^2(3x+1−2x+1)}{(3x+1{)}^3}\\ f\text{'}& \left(x\right)=\frac{6(2x−1{)}^2(x+2)}{(3x+1{)}^3}\end{array} \end{gathered}$$

To find the derivative we will use the product rule of differentiation.

So,

$$\begin{gathered} f\left(x\right)=3x^2{e}^{-4x} \\ \begin{array}{rl}f\text{'}& \left(x\right)=3(e^{−4x}\frac{d}{dx}{x}^2+x^2\frac{d}{dx}{e}^{−4x})\\ f\text{'}& \left(x\right)=3\left(e^{−4x}\right(2x)+x^2{e}^{−4x}\frac{d}{dx}(−4x\left)\right)\\ f\text{'}& \left(x\right)=3e^{−4x}(2x−4x^2)\end{array} \end{gathered}$$

To find the derivative we will use the chain rule of differentiation.

So,

$$\begin{gathered} f\left(x\right)=\sin \left(\ln x\right) \\ f\text{'}\left(x\right)=\cos \left(\ln x\right)\frac{d}{dx}\left(\ln x\right) \\ f\text{'}\left(x\right)=\frac{\cos \left(\ln x\right)}{x} \end{gathered}$$