Q. 1 - Calculus

Question

Differentiation: Find the derivative of each function f, and then simplify as much as possible.

$$\begin{gathered} \left(a\right) f\left(x\right)={\left(2x-1\right)}^3{\left(3x+1\right)}^2 \\ \left(b\right) f\left(x\right)=\frac{{\left(2x-1\right)}^3}{{\left(3x+1\right)}^2} \\ \left(c\right) f\left(x\right)=3x^2{e}^{-4x} \\ \left(d\right) f\left(x\right)=\sin \left(\ln x\right) \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The derivative of the given function is $f\text{'}\left(x\right)=30x(3x+1)(2x-1{)}^2.$

Part (b) The derivative of the given function is $f\text{'}\left(x\right)=\frac{6(2x-1{)}^2(x+2)}{(3x+1{)}^3}.$

Part (c) The derivative of the given function is $f\text{'}\left(x\right)=3e^{-4x}(2x-4x^2).$

Part (d) The derivative of the given function is $f\text{'}\left(x\right)=\frac{\cos \left(\ln x\right)}{x}.$

Part (a) Step 1. Given Information.

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

Part (a) Step 2. Find the derivative.

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

So,

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

Part (b) Step 1. Find the derivative.

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

So,

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

Part (c) Step 1. Find the derivative.

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

So,

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

Part (d) Step 1. Find the derivative.

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

So,

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