Question

using the definition of derivative, find the derivative of f(x)=2x^2 + 3x.
4x + 3
4h + 3
4x + 4h + 3
3x + 3h + 4
3x + 4
question 13 1 pts
using the definition of derivative, find the derivative of f(x)=\sqrt{2x + 1}.
\sqrt{2x + 1}
\frac{2}{\sqrt{2x+2h + 1}}
-2\sqrt{2x + 1}
\frac{2}{\sqrt{2x+1}}
\frac{1}{\sqrt{2x+1}}

Explanation:

Step1: Recall derivative definition

The definition of the derivative of a function $y = f(x)$ is $f^\prime(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$.

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 $h
ightarrow0$, $\lim_{h
ightarrow0}(4x + 2h+3)=4x+3$.

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 $h
ightarrow0$, $\lim_{h
ightarrow0}\frac{2}{\sqrt{2x+2h + 1}+\sqrt{2x + 1}}=\frac{1}{\sqrt{2x + 1}}$.

Answer:

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