1. Find using the definition........the gradient function f '(x)

Find using the defintion f '(c) lim = f(x) - f(c)
x -> c _______
x - c

the gradient function f ' (x) for:

a) f(x) = 1/x

b)f(x) =
√ x

[My tutor told me to rationalize before doing the question for part b but how do i rationalize this]

2. Hi
Originally Posted by sweetG
[My tutor told me to rationalize before doing the question for part b but how do i rationalize this]
To rationalize the numerator of $\frac{\sqrt{x}-\sqrt{c}}{x-c}$, multiply and divide the fraction by the conjugate of $\sqrt{x}-\sqrt{c}$ which is $\sqrt{x}+\sqrt{c}$ :

$\frac{\sqrt{x}-\sqrt{c}}{x-c}=\frac{(\sqrt{x}-\sqrt{c})(\sqrt{x}+\sqrt{c})}{(x-c)(\sqrt{x}+\sqrt{c})}=\ldots$

(notice that the numerator looks like $(a-b)(a+b)$ which equals $a^2-b^2$)

Edit : for the first question : $\frac{f(x)-f(c)}{x-c}=\frac{\frac{1}{x}-\frac{1}{c}}{x-c}=\frac{\frac{c-x}{xc}}{x-c}=\frac{c-x}{xc(x-c)}=-\frac{1}{xc}$ hence $f'(c) = ?$

3. Sorry I didn't realize that the definition for f '(c) was all over the place....sorry about that.

For Part b is the answer:

(x-c)/(x
√x + x√c - c√x - c√c

and for Part (a) how am I supposed to find f '(x)
Sorry this question is really confusing me
coz there are so many x's and c's.......

4. Originally Posted by sweetG
[SIZE=1][SIZE=2]
For Part b is the answer:

(x-c)/(x√x + x√c - c√x - c√c)
That's it but if you try computing $f'(c)=\lim_{x\to c}\frac{x-c}{x\sqrt{x}+x\sqrt{c}-c\sqrt{x}-c\sqrt{c}}$ you'll see that this limit is an indeterminate form. (both the numerator and the denominator tend to 0) If you hadn't expanded the numerator, it would have given you

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

And $f'(c)=\lim_{x\to c}\frac{1}{\sqrt{x}+\sqrt{c}}$ isn't an indeterminate form any longer. Can you find the value of this limit ?

and for Part (a) how am I supposed to find f '(x)
Sorry this question is really confusing me
coz there are so many x's and c's.......
The gradient of $f$ taken at $c$ is $f'(c)=\lim_{x\to c} -\frac{1}{xc}= -\frac{1}{c\cdot c}=-\frac{1}{c^2}$ as long as $c\neq 0$. Thus $\boxed{f'(x)=-\frac{1}{x^2}}$ for $x\neq 0$.

5. Does f '(c) = 1/2√c

which therefore makes f ' (x) = 1/2√x

Is this right or have I made a mistake ?

6. Originally Posted by sweetG
Does f '(c) = 1/2√c

which therefore makes f ' (x) = 1/2√x
That's it !