Derivative as a Limit: Finding dy/dx for y = √(2x+3)

[Karol]

Homework Statement

Use the definition of the derivative to find dy/dx for $~y=\sqrt{2x+3}$

Homework Equations

Derivative as a limit:
$$y'=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

The Attempt at a Solution

$$\lim_{\Delta x\rightarrow 0}\frac{\sqrt{2(x+\Delta x)+3}-\sqrt{2x+3}}{\Delta x}$$
$$=\lim_{\Delta x\rightarrow 0}\frac{\sqrt{2x+3+\Delta x}-\sqrt{2x+3}}{\Delta x}$$
The nominator and denominator tend to 0 but i can't find the limit

 

[blue_leaf77]

Have you tried making use of the relation $(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}) = a-b$?

$$=\lim_{\Delta x\rightarrow 0}\frac{\sqrt{2x+3+\Delta x}-\sqrt{2x+3}}{\Delta x}$$

Check again whether you miss any factor in front of $\Delta x$ in the numerator.

 

[Ray Vickson]

Homework Statement

Use the definition of the derivative to find dy/dx for $~y=\sqrt{2x+3}$

Homework Equations

Derivative as a limit:
$$y'=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

The Attempt at a Solution

$$\lim_{\Delta x\rightarrow 0}\frac{\sqrt{2(x+\Delta x)+3}-\sqrt{2x+3}}{\Delta x}$$
$$=\lim_{\Delta x\rightarrow 0}\frac{\sqrt{2x+3+\Delta x}-\sqrt{2x+3}}{\Delta x}$$
The nominator and denominator tend to 0 but i can't find the limit

$$\sqrt{2x+3+2 \Delta x} = \sqrt{2x+3} \sqrt{1 + \frac{2 \Delta x}{2x+3}}.$$
Can you find the behavior of $\sqrt{1 + h}$ or $\sqrt{1+h} - 1$ for small $h = 2 \Delta x/(2x+3)$?

 

[Karol]

$$\frac{\sqrt{2(x+\Delta x)+3}-\sqrt{2x+3}}{\Delta x}=\frac{\sqrt{2(x+\Delta x)+3}-\sqrt{2x+3}}{\Delta x} \cdot \frac{\sqrt{2(x+\Delta x)+3}+\sqrt{2x+3}}{\sqrt{2(x+\Delta x)+3}+\sqrt{2x+3}}=\frac{2\Delta x}{\Delta x(\sqrt{2(x+\Delta x)+3}+\sqrt{2x+3})}=\frac{1}{\sqrt{2x+3}}$$

Can you find the behavior of $\sqrt{1 + h}$ or $\sqrt{1+h} - 1$ for small $h = 2 \Delta x/(2x+3)$?

$$\lim_{h\rightarrow 0}\sqrt{1+h}=1$$
$$=\lim_{\Delta x\rightarrow 0}\frac{\sqrt{2x+3+2\Delta x}-\sqrt{2x+3}}{\Delta x}=\frac{\sqrt{2x+3}-\sqrt{2x+3}}{\Delta x}=\frac{0}{0}$$

 

[Ray Vickson]

$$\frac{\sqrt{2(x+\Delta x)+3}-\sqrt{2x+3}}{\Delta x}=\frac{\sqrt{2(x+\Delta x)+3}-\sqrt{2x+3}}{\Delta x} \cdot \frac{\sqrt{2(x+\Delta x)+3}+\sqrt{2x+3}}{\sqrt{2(x+\Delta x)+3}+\sqrt{2x+3}}=\frac{2\Delta x}{\Delta x(\sqrt{2(x+\Delta x)+3}+\sqrt{2x+3})}=\frac{1}{\sqrt{2x+3}}$$

$$\lim_{h\rightarrow 0}\sqrt{1+h}=1$$
$$=\lim_{\Delta x\rightarrow 0}\frac{\sqrt{2x+3+2\Delta x}-\sqrt{2x+3}}{\Delta x}=\frac{\sqrt{2x+3}-\sqrt{2x+3}}{\Delta x}=\frac{0}{0}$$

Nope: that answer is of no use. By "behavior" for small $h$ I mean: how does $\sqrt{1+h}-1$ approach 0? Does it go to zero like $a h^{10}$ ($a$ = constant)? Does it go to zero like $b h$ ($b$ = constant)? Does it go to zero like $c \sqrt{h}?$ For small $h$ you need to get some simple function $f(h) \to 0$ in the numerator, so you can see what the ratio $f(h)/h$ looks like for small but non-zero $h$.

 

[Karol]

$$\frac{\sqrt{2x+3+2\Delta x}-\sqrt{2x+3}}{\Delta x}=\frac{ \sqrt{2x+3} \sqrt{1 + \frac{2 \Delta x}{2x+3}}-\sqrt{2x+3}}{\Delta x}=\frac{\sqrt{2x+3}\left[ \sqrt{1+\frac{2\Delta x}{2x+3}}-1 \right]}{\Delta x}$$
$$=\frac{\sqrt{2x+3}\left[ \sqrt{1+\frac{2\Delta x}{2x+3}}-1 \right]}{\Delta x} \cdot \frac{ \sqrt{1+\frac{2\Delta x}{2x+3}}+1 }{ \sqrt{1+\frac{2\Delta x}{2x+3}}+1 }$$
$$=\frac{1}{\sqrt{2x+3}}$$

 

[Ray Vickson]

$$\frac{\sqrt{2x+3+2\Delta x}-\sqrt{2x+3}}{\Delta x}=\frac{ \sqrt{2x+3} \sqrt{1 + \frac{2 \Delta x}{2x+3}}-\sqrt{2x+3}}{\Delta x}=\frac{\sqrt{2x+3}\left[ \sqrt{1+\frac{2\Delta x}{2x+3}}-1 \right]}{\Delta x}$$
$$=\frac{\sqrt{2x+3}\left[ \sqrt{1+\frac{2\Delta x}{2x+3}}-1 \right]}{\Delta x} \cdot \frac{ \sqrt{1+\frac{2\Delta x}{2x+3}}+1 }{ \sqrt{1+\frac{2\Delta x}{2x+3}}+1 }$$
$$=\frac{1}{\sqrt{2x+3}}$$

Alternatively:
$$\sqrt{1+h}-1 = \frac{(\sqrt{h+1}-1)(\sqrt{h+1}+1)}{\sqrt{h+1}+1} = \frac{(h+1)-1}{\sqrt{h+1}+1} \approx \frac{h}{2}$$
for small $|h|$. Thus,
$$f(x+\Delta x)-f(x) \approx \sqrt{2x+3}\frac{1}{2 (2x+3)} 2 \Delta x = \frac{\Delta x}{\sqrt{2x+3}}$$
so $f'(x) = 1/\sqrt{2x+3}.$

 

[Karol]

Thank you Ray