Answer:
a) [tex]\displaystyle{\dfrac{dG}{dV_{\text out}}=\dfrac{20}{V_{\text out}\ln 10}}[/tex]
b) [tex]\displaystyle{\dfrac{d^2G}{dV_\text{out}^2} = -\dfrac{20}{V_\text{out}^2\ln 10}}[/tex]
Step-by-step explanation:
From the given equations to find, we are differentiating function G with respect to V out. Therefore, we will be only using the equation (6):
[tex]\displaystyle{G=20\log (10V_{\text out})}[/tex]
Here’s a fresh-up reminders for logarithmic differentiation formula:
[tex]\displaystyle{f(V)=a\log_b V \to f'(x)=\dfrac{a\cdot V'}{V\ln b}}[/tex]
Note that V' means to derive or differentiate the V-term
Therefore, in this scenario, our values of term are:
- a = 20
- b = 10 (Since logarithm base is not shown, it’s always assumed to be common logarithm)
- V = 10V out
Therefore, we can apply differentiation formula:
[tex]\displaystyle{\dfrac{dG}{dV_{\text out}}=\dfrac{20\cdot \dfrac{d(10V_{\text out})}{dV_{\text out}}}{10V_{\text out}\ln 10}}[/tex]
Simplify:
[tex]\displaystyle{\dfrac{dG}{dV_{\text out}}=\dfrac{20\cdot 10}{10V_{\text out}\ln 10}}\\\\\displaystyle{\dfrac{dG}{dV_{\text out}}=\dfrac{20}{V_{\text out}\ln 10}}[/tex]
And we have finished the part a. On part b, we just differentiate the answer from part a again, the second part is to obtain the second derivative which is to derive the first derivative. Therefore:
[tex]\displaystyle{\dfrac{d^2G}{dV_{\text out}^2} = \dfrac{d\left(\dfrac{20}{V_{\text out}\ln 10}\right)}{dV_{\text out}}}[/tex]
First, separate the constant every time when you have to differentiate a function:
[tex]\displaystyle{\dfrac{20}{\ln 10}\cdot \dfrac{1}{V_{\text out}}}[/tex]
Differentiate 1/V out: recall the formula of fraction:
[tex]\displaystyle{f(V)=\dfrac{1}{V} =V^{-1} \to f'(x)=-V^{-2} = -\dfrac{1}{V^2}[/tex]
The formula above is derived from power rules formula where:
[tex]\displaystyle{f(V)=V^n = nV^{n-1}}[/tex]
Therefore, from the function of V out:
[tex]\displaystyle{\dfrac{d^2G}{dV_\text{out}^2} = \dfrac{20}{\ln 10} \cdot \left(\dfrac{1}{V_\text{out}}\right)'}\\\\\displaystyle{\dfrac{d^2G}{dV_\text{out}^2} = \dfrac{20}{\ln 10} \cdot (V_\text{out})^{-1}}\\\\\displaystyle{\dfrac{d^2G}{dV_\text{out}^2} = \dfrac{20}{\ln 10} \cdot (-V_\text{out})^{-2}}\\\\\displaystyle{\dfrac{d^2G}{dV_\text{out}^2} = \dfrac{20}{\ln 10} \cdot \left(-\dfrac{1}{V_\text{out}^2}\right)}[/tex]
Simplify to latest as we get:
[tex]\displaystyle{\dfrac{d^2G}{dV_\text{out}^2} = -\dfrac{20}{V_\text{out}^2\ln 10}}[/tex]
Please let me know if you have any questions!
