Question

Find the slope of the tangent to the curve at the given point.

Answer 1

\[y=\frac{ 8 }{ \sqrt{x+11} }\] at x=5

Answer 2

Are you in a calculus class?

Answer 3

Yep.

Answer 4

Do you know how to take the derivative of that function?

Answer 5

The first derivative of the function gives you a formula for the instantaneous slope of the function. Plug in your value of \(x\) and you've got the slope of the tangent to the curve at that point.

Answer 6

I tried but didn't get the right answer.

Answer 7

How did you do it?

Answer 8

Using limiting but I don't think the formula that I got is right.

Answer 9

Well, you can use the chain rule and power rule.

if we let \(u = x+11\) then we can rewrite it as \[y = 8u^{-1/2}\]

Answer 10

derivative of \[y = 8u^{-1/2}\]is\[-\frac{1}{2}*8u^{-3/2} * \frac{du}{dx} = -4(x+11)^{-3/2} = -\frac{4}{\sqrt{(x+11)^3}}\]

Answer 11

if \(u = x+11\)\[\frac{du}{dx} = 1\]