using the definition of derivative, find the derivative of f(x)=2x^2 + 3x.\n4x + 3\n4h + 3\n4x + 4h + 3\n3x…

Explanation:

Step1: Recall derivative definition

The definition of the derivative of a function $y = f(x)$ is $f^\prime(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$.

Step2: For $f(x)=2x^{2}+3x$

First, find $f(x + h)$: [ \begin{align*} f(x + h)&=2(x + h)^{2}+3(x + h)\ &=2(x^{2}+2xh+h^{2})+3x + 3h\ &=2x^{2}+4xh+2h^{2}+3x + 3h \end{align*} ] Then, $\frac{f(x + h)-f(x)}{h}=\frac{(2x^{2}+4xh+2h^{2}+3x + 3h)-(2x^{2}+3x)}{h}=\frac{4xh+2h^{2}+3h}{h}=4x + 2h+3$. Taking the limit as $h\rightarrow0$, $\lim_{h\rightarrow0}(4x + 2h+3)=4x+3$.

Step3: For $f(x)=\sqrt{2x + 1}$

Find $f(x + h)$: $f(x + h)=\sqrt{2(x + h)+1}=\sqrt{2x+2h + 1}$. [ \begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{\sqrt{2x+2h + 1}-\sqrt{2x + 1}}{h}\ &=\frac{(\sqrt{2x+2h + 1}-\sqrt{2x + 1})(\sqrt{2x+2h + 1}+\sqrt{2x + 1})}{h(\sqrt{2x+2h + 1}+\sqrt{2x + 1})}\ &=\frac{(2x+2h + 1)-(2x + 1)}{h(\sqrt{2x+2h + 1}+\sqrt{2x + 1})}\ &=\frac{2h}{h(\sqrt{2x+2h + 1}+\sqrt{2x + 1})}\ &=\frac{2}{\sqrt{2x+2h + 1}+\sqrt{2x + 1}} \end{align*} ] Taking the limit as $h\rightarrow0$, $\lim_{h\rightarrow0}\frac{2}{\sqrt{2x+2h + 1}+\sqrt{2x + 1}}=\frac{1}{\sqrt{2x + 1}}$.

Answer:

Question 12: A. $4x + 3$ Question 13: E. $\frac{1}{\sqrt{2x + 1}}$