Can someone help me find derivative/tangent line?

Question

anonymous · Answer

I found the derivative of $$\sqrt{x}$$ to be $$\frac{ 1 }{ 2x ^{1/2} }$$

Then, when asked to find the tangent line at (9,3), I got the equation: y=1/6x+3/2. I was then asked to find all values of x such that the slope is 1/4. 

Am I doing the derivative/tangent line right? and if so, how would I compute all values of x such that the slope is 1/4?

anonymous · Answer

http://www.dummies.com/how-to/content/how-to-find-the-derivative-of-a-line.html

anonymous · Answer

check that out. it even helped me

anonymous · Answer

Thanks, will check it out!

anonymous · Answer

np :)

anonymous · Answer

I calculated the value 4 as a x-value with a slope of 1/4... would that work?

phi · Answer

yes.  you want m= 1/4.  you found
$$ \text{slope } m = \frac{1}{2 \sqrt{x}} \ 
 \frac{1}{2 \sqrt{x}}= \frac{1}{4} \
2 \sqrt{x}= 4\
\sqrt{x}= 2\
x = 2^2 = 4
$$

phi · Answer

Here is a graph of y= sqr(x)  and the tangent line (with slope = 0.25 = 1/4)

anonymous · Answer

Thank you so much!