Derivative

[volc]

Find the y-intercept of the tangent line to the graph of y = (9)/sqrt(1 + x) at the point where x = 9.

I used the quotient rule to find the derivative.

I found that to be (-9/2(1+x)^-1/2)/(1+x) But then I am totally clued out...

Thanks for helping me out.

volc

[Physics Monkey]

What is the relationship between the derivative and the tangent line?

[hotvette]

You need to assess what you have:

1. slope of the line you want to construct
2. a point that the line goes through

These 2 pieces of information should allow you to write the equation of the line. Once you have the equation, you can find the y intercept.

[volc]

Derivative of function = slope of the tgt line.

[volc]

I don't understand what you mean. :/

[Physics Monkey]

If you have the equation of the tangent line, you can find the y-intercept, right? You need therefore to find the equation of the line. You have two pieces of information, the slope of the line and a point the line passes through. Is this suffucient to write an equation for the line?

[amcavoy]

I don't understand what you mean. :/
A line in the x-y plane is given by y=mx+b (I assume you know that). Taking the derivative and plugging in your x=9 will give you m. You still need to find b, which is precisely what the question asks for. If I were you, I would plug x=9 into your original equation, find your y value, then substitute both of those into y=mx+b and solve for b.

Alex

[volc]

(-9/2(1+x)^-1/2)/(1+x) = slope of the tgt line right ? And the equation for the tangental line is y= mx + b, where m = (-9/2(1+x)^-1/2)/(1+x). To find y I need to find b and I can't seen to be able to find b.

[amcavoy]

(-9/2(1+x)^-1/2)/(1+x) = slope of the tgt line right ? And the equation for the tangental line is y= mx + b, where m = (-9/2(1+x)^-1/2)/(1+x). To find y I need to find b and I can't seen to be able to find b.
That is m(x). To find the slope at a point, you need to plug that x-value in!:smile: Do you understand why?

Alex

[volc]

When I plug in the x value (-9/2(1+x)^-1/2)/(1+x) I get:
(-9/2)((10)^-3/2) and this is x or m(x) ?

volc

[amcavoy]

When I plug in the x value (-9/2(1+x)^-1/2)/(1+x) I get:
(-9/2)((10)^-3/2) and this is x or m(x) ?

volc
What do you think it is? Remember that the derivative at a point will be the slope (m) of the graph at that point. Now, you have an x-value given to you. You need to find the cooresponding y-value so that you have a slope and a point. You can find an equation of a line using a point and the slope, thus find the y-intercept.

If I were you, I'd go back and read up on what a derivative actually means geometrically.

Alex