# Find the slope of the tangent line

• Oct 11th 2009, 01:40 PM
DBA
Find the slope of the tangent line
hi, I did something wrong but I do not what:
y = f(x)=1/sqrt(x) = x^-1/2

slope m= lim h->0 {f(x+h) - f(x)]/h

m=f'(x)= lim h->0 [1/sqrt(x+h) - 1/sqrt(x)]/h

then I multiply by [sqrt(x+h) + sqrt(x)] / [sqrt(x+h) + sqrt(x)]

m= lim h->0 {[x-(x+h)]/[sqrt(x+h)*sqrt(x)]} / h

m= lim h->0 h / [h*(sqrt(x+h)*sqrt(x)] = -1/2sqrt(x)

This is wrong, it must be -1/ 2sqrt(x^3)

Thanks for any help
• Oct 11th 2009, 01:54 PM
pickslides
Quote:

Originally Posted by DBA
hi, I did something wrong but I do not what:
y = f(x)=1/sqrt(x) = x^-1/2

$\displaystyle f(x)= x^{\frac{-1}{2}}$

First principles is very messy here, using $\displaystyle y = x^n \Rightarrow y' = nx^{n-1}$

$\displaystyle f'(x)= \frac{-1}{2}x^{\frac{-3}{2}}$
• Oct 11th 2009, 02:05 PM
DBA
Thank you for the reply, but I need to use the formula to solve the problem
slope m= lim h->0 [f(x+h) - f(x)]/h

for f(x) = 1/sqrt(x)

I know the answer, I just do not know how to get there with this formula.

Thanks