# Thread: Derivative using limit definition and line tangent to graph

1. ## Derivative using limit definition and line tangent to graph

Hi!

Apparently I made mistakes when solving this, could I be helped in spotting the errors please?

a. Given $f(x)=\sqrt{x-2}+1$, find $f^{\prime}(x)$ using the limit definition of the derivative.

$f^{\prime}(x) = \frac1h \cdot \frac{h+2}{\left(\sqrt{x+h-2}+1+\sqrt{x-2}+1\right)}$

$= \frac{2}{\left(\sqrt{x+h-2}+1+\sqrt{x-2}+1\right)}$

$= \frac{2}{2\sqrt{x-2}+2}$

$= \frac{1}{\sqrt{x-2}+2}$

b. Find the equation of the line tangent to the graph $f(x)=\sqrt{x-2}+1$ at the point x = 6.

$y = f(x)(f^{\prime}(x)(x-x_0)$

$f^{\prime}(x) = \frac{1}{2+2} = \frac14$

$f(x) = \left(\sqrt{6-2}+1\; = 3\right)$

$y = 3\cdot\frac14(x-6)$

2. ## Re: Derivative using limit definition and line tangent to graph

Originally Posted by Unreal
Hi!

Apparently I made mistakes when solving this, could I be helped in spotting the errors please?

a. Given $f(x)=\sqrt{x-2}+1$, find $f^{\prime}(x)$ using the limit definition of the derivative.
rather than try and spot your error I'm just going to go through how it should be done.

$\frac{d}{dx}f[x]=\text{limit}_{h\rightarrow0}\frac{f[x+h] - f[x]}{h}$

$\frac{d}{dx}f[x]=\text{limit}_{h\rightarrow0}\frac{(\sqrt{x+h-2}+1)-(\sqrt{x-2}+1)}{h}$

$\frac{d}{dx}f[x]=\text{limit}_{h\rightarrow0}\frac{\sqrt{x+h-2}-\sqrt{x-2}}{h}$

$\frac{d}{dx}f[x]=\text{limit}_{h\rightarrow0}\frac{(x+h-2)-(x-2)}{h(\sqrt{x+h-2}+\sqrt{x-2})}$

$\frac{d}{dx}f[x]=\text{limit}_{h\rightarrow0}\frac{h}{h(\sqrt{x+h-2}+\sqrt{x-2})}$

$\frac{d}{dx}f[x]=\text{limit}_{h\rightarrow0}\frac{1}{(\sqrt{x+h-2}+\sqrt{x-2})}$

$\frac{d}{dx}f[x]=\frac{1}{2\sqrt{x-2}}$

ok here's this for now so you can look at it. You actually made only one small mistake in keeping track of a negative sign.

3. ## Re: Derivative using limit definition and line tangent to graph

Originally Posted by Unreal
Hi!

Apparently I made mistakes when solving this, could I be helped in spotting the errors please?

a. Given $f(x)=\sqrt{x-2}+1$, find $f^{\prime}(x)$ using the limit definition of the derivative.

$f^{\prime}(x) = \frac1h \cdot \frac{h+2}{\left(\sqrt{x+h-2}+1+\sqrt{x-2}+1\right)}$
The "difference quotient" would be $\frac{(\sqrt{x+h- 2}+ 1)- (\sqrt{x-2}+1)}{h}= \frac{\sqrt{x+h- 2}- \sqrt{x- 2}}{h}$
Now multiplying numerator and denominator by $\sqrt{x+h- 2}+ \sqrt{x- 2}$ we have $\frac{x+ h- 2- (x- 2)}{h(\sqrt{x+h- 2}+ \sqrt{x- 2})}= \frac{h}{h(\sqrt{x+h- 2)+ \sqrt{x- 2})}$. That is, there is no "+ 2" in the numerator.

$= \frac{2}{\left(\sqrt{x+h-2}+1+\sqrt{x-2}+1\right)}$

$= \frac{2}{2\sqrt{x-2}+2}$

$= \frac{1}{\sqrt{x-2}+2}$

b. Find the equation of the line tangent to the graph $f(x)=\sqrt{x-2}+1$ at the point x = 6.

$y = f(x)(f^{\prime}(x)(x-x_0)$

$f^{\prime}(x) = \frac{1}{2+2} = \frac14$

$f(x) = \left(\sqrt{6-2}+1\; = 3\right)$

$y = 3\cdot\frac14(x-6)$