Find an equation of the line tangent to the graph of 6/sqrtx at the point (25,6/5)

please help?

Question

Find an equation of the line tangent to the graph of 6/sqrtx at the point (25,6/5) 

please help?

mertsj · Answer

Find the first derivative.

anonymous · Answer

I did find the first derivative then I plugged 25 into the f'(x) which will give me the slope.  But that was wrong.  I used y-y1 = m(x-x1)

mertsj · Answer

What did you get for the first derivative?

anonymous · Answer

after finding derivative whether by the quotient rule or just by re-organizing the equation, I get f'(x) = 6 * (-1/2)x^(-3/2)

then, at x = 25, I got 0.024 = which is slope.

y - y 1 = m (x-x1)

I plugged 6/5 for y 1 = 0.024 for m and 25 for x1

anonymous · Answer

I got the equation 0024x+06 but still wrong?

mertsj · Answer

Probably should leave the slope in its exact form.

mertsj · Answer

Also, how is the slope not negative?

mertsj · Answer

$$f'(x)=\frac{-3}{x ^{\frac{3}{3}}}$$

anonymous · Answer

Wait let me try it again.

anonymous · Answer

Yea I got it right now because I didn't have minus

anonymous · Answer

Thanks!

mertsj · Answer

yw