Question

Pre calculus homework help

Answer 1

`f'((1)/(3√(e)))=(1)/(2)`

Explanation:

Find f'(x) using Product Rule

`f(x)=x\ln(3x)\\f'(x)=\ln(3x)+3x((1)/(3x))\\f'(x)=\ln(3x)+1\\\\f'((1)/(3√(e)))=\ln(3\cdot(1)/(3√(e)))+1\\\\f'((1)/(3√(e)))=\ln((1)/(√(e)))+1\\\\f'((1)/(3√(e)))=\ln(e^{-(1)/(2)})+1\\\\f'((1)/(3√(e)))=-(1)/(2)\ln(e)+1\\\\f'((1)/(3√(e)))=-(1)/(2)+1\\\\f'((1)/(3√(e)))=(1)/(2)`

Answer 2

`f'\left((1)/(3√(e))\right)=(1)/(2)`

Explanation:

Given function:

`f(x)=x\ln(3x)`

To find f'(x), differentiate the given function using the product rule.

`\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\frac{\text{d}y}{\text{d}x}=u\frac{\text{d}v}{\text{d}x}+v\frac{\text{d}u}{\text{d}x}$\\\end{minipage}}`

`\textsf{Let\;$u=x^2}`
`\textsf{Let\;$u=x$}\implies \frac{\text{d}u}{\text{d}x}=1`

`\textsf{Let\;$v=\ln(3x)$}\implies \frac{\text{d}v}{\text{d}x}=(1)/(3x)\cdot 3=(1)/(x)`

Input the values into the product rule to differentiate the function:

`\begin{aligned}\frac{\text{d}y}{\text{d}x}&=u\frac{\text{d}v}{\text{d}x}+v\frac{\text{d}u}{\text{d}x}\\\\&=x \cdot (1)/(x)+\ln(3x) \cdot 1\\\\&=1+\ln(3x)\end{aligned}`

To find the value of f'(1/(3√e)), substitute x = 1/(3√e) into the differentiated function:

`\begin{aligned}f'\left((1)/(3√(e))\right)&=1+\ln\left(3\left((1)/(3√(e))\right)\right)\\\\&=1+\ln\left((1)/(√(e))\right)\\\\&=1+\ln e^{-(1)/(2)}\\\\&=1-(1)/(2)\ln e\\\\&=1-(1)/(2)(1)\\\\&=1-(1)/(2)\\\\&=(1)/(2)\end{aligned}`

`\hrulefill`

Differentiation rules used:

`\boxed{\begin{minipage}{4 cm}\underline{Differentiating $ax$}\\\\If $y=ax$, then $\frac{\text{d}y}{\text{d}x}=a$\\\end{minipage}}`

`\boxed{\begin{minipage}{6 cm}\underline{Differentiating $\ln(f(x))$}\\\\If $y=\ln(f(x))$, then $\frac{\text{d}y}{\text{d}x}=(1)/(f(x))\cdot f'(x)$\\\end{minipage}}`