Derivatives

[KidInkFan]

Hi, just need help starting the following question:

Find an equation of the tangent line to the the graph of the function when x=9
f(x) = x^(1/2)(10-6x^2)/(x) i already put in the value where x = 9 and got 158.666 but not i have to compute f'(x). Do I expand the x^(1/2) and then take the derivative?
Does this involve the quotient rule? I tried it out 1(10^(1/2)-6x)-x((1/2*10)x^(-1/2)-6)/(x)^2 i hope i applied the brackets correctly

Thank-you for your help!!

 

[DrPhil]

Hi, just need help starting the following question:

Find an equation of the tangent line to the the graph of the function when x=9
f(x) = x^(1/2)(10-6x^2)/(x) i already put in the value where x = 9 and got 158.666 but not i have to compute f'(x). Do I expand the x^(1/2) and then take the derivative?
Does this involve the quotient rule? I tried it out 1(10^(1/2)-6x)-x((1/2*10)x^(-1/2)-6)/(x)^2 i hope i applied the brackets correctly

Thank-you for your help!!

Since the denominator is simply x, I would do the division and separate the two terms:

\(\displaystyle \displaystyle f(x) = 10\ x^{-1/2} - 6\ x^{3/2}\)

That should be easier to differentiate - anything that avoids the quotient rule helps. In your result, if you do the division by x^2 and combine terms, you should get the same answer.

I am glad you recognize that you differentiate first, and then evaluate at x=9 last to find the slope of the tangent line.

 

[lookagain]

Find an equation of the tangent line to the the graph of the function when x=9
f(x) = x^(1/2)(10-6x^2)/(x) i already put in the value where x = 9 and got > > > 158.666 < < <
The function value at x = 9 is not positive.

If your x-value and function have been typed correctly, then \(\displaystyle f(9) \approx -158.667\)