# Equation of tangent line at sqrt(x)

1. Oct 14, 2006

Hey, right now in our calculus course we are currently working on "derivatives" using the differentiation equations. I was hoping that someone could help me out with the following question:

Find the equation of the tangent line to the curve at the given point.

y = rootX , (4, 0.4)
------
X + 1

It's not an overly difficult question because we know that the equation is y-0.4 = m(x-4) and we are looking for m(the slope), and we also know that we should be using the formula f'(x)g(x)-g(x)f'(x)/[g(x)]^2

However, I'm still having some difficulties in working this out, if someone could please shed some light on this, I'd be happy. :D

2. Oct 14, 2006

So $$y = \frac{\sqrt{x}}{x+1}$$. Have you used the quotient rule?

$$\frac{dy}{dx} = \frac{(x+1)\frac{1}{2}x^{-\frac{1}{2}} - x^{\frac{1}{2}}}{(x+1)^{2}}$$

Can you simplify this? You could just put in x = 4, without simplifying. Do you understand how to get the above expression?

$$f(x) = \sqrt{x}$$ and $$g(x) = x+1$$

So what is $$f'(x)$$ and $$g'(x)$$?

Last edited: Oct 14, 2006
3. Oct 14, 2006

### cepheid

Staff Emeritus
Well, you have the method, so now work it out step by step:

$$f(x) = \sqrt{x}$$

$$g(x) = x+1$$

$$\left(\frac{f(x)}{g(x)}\right)' = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}$$

Anyway, this is your cue to take over:

$$f'(x) = \textrm{?}$$

$$g'(x) = \textrm{?}$$

Substitute these expressions into the expression for the derivative and evaluate at x = 4.

Last edited: Oct 14, 2006
4. Oct 14, 2006

Ahhh alrighty, well I understand what you are saying, but I did the following:
dy
-- = (x+1)d/dx (squarerootx) - (squarerootx)d/dx(x+1)
dx -------------------------------
(X+1)^2

= (x+1)1 - (squarerootX)1
-------
2squarerootX
---------------------------------
(x+1)^2

Is that stating the same thing as you had, or have I misinterpreted this.

Also, after that step, I'm not sure how to simplify it though. I know the (x+1)^2 moves up and the denominator becomes 2squarerootX(X+1)^2 but it's solving in the numerator that is giving me troubles.

Last edited: Oct 14, 2006
5. Oct 14, 2006

Actually nevermind what I just said, how abouts do you simplify the expression you just gave me courtrigad? Could you just take out the x^-0.5?

6. Oct 14, 2006

$$\frac{dy}{dx} = \frac{(x+1)\frac{1}{2}x^{-\frac{1}{2}} - x^{\frac{1}{2}}}{(x+1)^{2}} = \frac{\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{1}}{x+1} = \frac{1-2x}{2\sqrt{x}(x+1)}$$

Divide by $$x+1$$ in the numerator and denominator and change $$x^{-\frac{1}{2}}$$ to $$\frac{1}{\sqrt{x}}$$

Last edited: Oct 14, 2006
7. Oct 14, 2006