Finding f'(4) for f(x) = x - sqrt(x)

[turophile]

For a tangent line problem, I need to find f'(4) for f(x) = x - sqrt(x). Here's what I have so far:

x = 4
y = f(x) = 4 - sqrt(4) = 4 - 2 = 2
f(4 + ?x) = 4 + ?x - sqrt(4 + ?x)
f(4 + ?x) - f(x) = 4 + ?x - sqrt(4 + ?x) - 2 = 2 + ?x - sqrt(4 + ?x)
[f(4 + ?x) - f(x)]/?x = [2 + ?x - sqrt(4 + ?x)]/?x

In the next step, I need to factor ?x out of the numerator so it will cancel out the ?x in the denominator (or do something else so the denominator will not approach 0 as ?x approaches 0). I'm having trouble seeing how to do this.

 

[Deleted member 4993]

For a tangent line problem, I need to find f'(4) for f(x) = x - sqrt(x). Here's what I have so far:

x = 4
y = f(x) = 4 - sqrt(4) = 4 - 2 = 2
f(4 + ?x) = 4 + ?x - sqrt(4 + ?x)
f(4 + ?x) - f(x) = 4 + ?x - sqrt(4 + ?x) - 2 = 2 + ?x - sqrt(4 + ?x)
[f(4 + ?x) - f(x)]/?x = [2 + ?x - sqrt(4 + ?x)]/?x

In the next step, I need to factor ?x out of the numerator so it will cancel out the ?x in the denominator (or do something else so the denominator will not approach 0 as ?x approaches 0). I'm having trouble seeing how to do this.

Find g'(x) for g(x) = x

Find h'(x) for h(x) = ?(x)

find f'(x) = g'(x) - h'(x)

then find f'(4)

 

[turophile]

I thought of that, but my textbook doesn't give "If h = f + g then h' = f' + g'" until the next section of the chapter on derivatives. In the section I'm working in now, we only have the ?-notation and definition of the derivative as ?y/?x to work with. I'm pretty sure I'm suppose to solve this one like the other slope problems in this section by just using the ?-notation and algebra.

Here's an example of another one that I did figure out successfully:

f(x) = x[sup:2qu6wtg2]2[/sup:2qu6wtg2] + 2x
x = - 1
y = f(- 1) = (- 1)[sup:2qu6wtg2]2[/sup:2qu6wtg2] + 2 (- 1) = 1 - 2 = - 1
f(- 1 + ?x) = (- 1 + ?x)[sup:2qu6wtg2]2[/sup:2qu6wtg2] + 2 (- 1 + ?x) = 1 - 2?x + (?x)[sup:2qu6wtg2]2[/sup:2qu6wtg2] - 2 - 2?x = - 1 + (?x)[sup:2qu6wtg2]2[/sup:2qu6wtg2]
?y = f(- 1 + ?x) - f(- 1) = - 1 + (?x)[sup:2qu6wtg2]2[/sup:2qu6wtg2] - (- 1) = (?x)[sup:2qu6wtg2]2[/sup:2qu6wtg2]
?y/?x = (?x)[sup:2qu6wtg2]2[/sup:2qu6wtg2]/?x = ?x
f'(- 1) = lim{x ? 0} ?y/?x = 0

So the slope of the line tangent to y = x[sup:2qu6wtg2]2[/sup:2qu6wtg2] + 2x at x = - 1 is 0, and the equation of the line is y - ( - 1 ) = 0 ( x - ( - 1 ) ) = 0 ? y = - 1.

My question is, can this method also be used to solve the problem I posted?

 

[galactus]

If I am understanding correctly, they want you to find the derivative using first principles or the definiton of a derivative.

\(\displaystyle f(x)=x-\sqrt{x}\)

It's a matter of hammering at it with some algebra. When confronted with radicals, try the conjugate thing.

\(\displaystyle \lim_{h\to 0}\frac{x+h-\sqrt{x+h}-(x-\sqrt{x})}{h}\)

\(\displaystyle \lim_{h\to 0}\frac{h-\sqrt{x+h}+\sqrt{x}}{h}\)

Multiply top and bottom by conjugate of \(\displaystyle \sqrt{x+h}-\sqrt{x}\)

\(\displaystyle \lim_{h\to 0}\left[1-\frac{(\sqrt{x+h}-\sqrt{x})}{h}\cdot\frac{(\sqrt{x+h}+\sqrt{x})}{(\sqrt{x+h}+\sqrt{x})}\right]\)

\(\displaystyle \lim_{h\to 0}\left[1-\frac{\not{h}}{\not{h}(\sqrt{x+h}+\sqrt{x})}\right]\)

Now, let h=0 and you should have the derivative of \(\displaystyle x-\sqrt{x}\). Plug in x=4 and that is the derivative at x=4 and, therefore, the slope of the tangent lone at that point.

Here is one with some meat on it. Using first principles, find the derivative of \(\displaystyle x^{\frac{-3}{4}}\)

Try giving it a go.

If you get stuck, look here: viewtopic.php?f=3&t=37962&p=147329&hilit=first+principles#p147329