1. ## Simple derivative problem

I've gotten the same answer for this very simple derivative problem countless times, and apparently it's wrong according to the site I'm doing my homework from. You are asked to find the derivative of 6/x and then the second derivative of this. I keep getting 6/x^2 for the first derivative...what could I be doing wrong?

2. Write it as: $f(x)=6x^{-1}$

Then $f'(x) = 6x^{-1-1} \cdot (-1)=-6x^{-2}=\frac{-6}{x^2}$

3. Wait how did you do that? I only know the lim as h approaches 0 f(x+h) etc etc formula for finding derivatives...

4. (x^n)' = n.x^(n-1)

Example:
(x^5)'= 5.x^4

5. wow...I wish my professor would have shown me that...haha

6. $f(x)= \frac{6}{x}$

$\lim_{h \to 0}~\frac{\frac{6}{x+h}-\frac{6}{x}}{h}$

$\lim_{h \to 0}~ \frac{\frac{6x-6x-6h}{x^2 +xh}}{h}$

$\lim_{h \to 0}~\frac{-6h}{h(x^2+xh)}$

$\lim_{h \to 0}~ \frac{-6}{x^2+xh}$

$f'(x) = \frac{-6}{x^2}$

7. Originally Posted by fattydq
wow...I wish my professor would have shown me that...haha
You will learn that after but first they teach you how to find the derivative

$\lim_{h \to 0}~\frac{f(x+h)-f(x)}{h}$

The other way dax918 and Spec showed is called the power rule

8. Also, you can find it using logaritmic differenciation, you'll learn that too soon.

y=6/x

Ln y = Ln 6 - Ln x

(Ln y)' = (Ln 6 - Ln x)'

(1/y)(dy/dx) = -1/x

(dy/dx) = (-1/x).(6/x) = -6/x^2

9. Originally Posted by dax918
(x^n)' = n.x^(n-1)

Example:
(x^5)'= 5.x^4
OK we actually did learn this formula today and I now have another question related to it. How would you use that formula to find the derivative (differentiate) of (7x^2+3x^1/2)/2x

10. Originally Posted by fattydq
OK we actually did learn this formula today and I now have another question related to it. How would you use that formula to find the derivative (differentiate) of (7x^2+3x^1/2)/2x
$\frac{7x^2+3\sqrt{x}}{2x}$

$\frac{7x^2}{2x} +\frac{3\sqrt{x}}{2x}$

$\frac{7}{2}x + \frac{3}{2}x^{\frac{-1}{2}}$

So now it should be simple to use the power rule