# Need help here with derivatives

• February 14th 2008, 08:13 AM
Need help here with derivatives
Hey guys. I'm having trouble with derivatives and I need someone's help with my homework. Could you guys please explain to me how to solve these problems?

1.
Given $y = f(x) = \frac{1}{x^2} - \sqrt[3]{x}$
Find the second derivative
$f"(x)=\frac{d^2y}{dx^2}$

2.
Given $f(x) = \frac{x^2}{1-3x}$
Find the derivative f'(x)

3.
Given $\Delta y = 2x^2(\Delta x+3x(\Delta x)^2-2(\Delta x)$
Find the derivative = $Lim \frac{f(x+\Delta x)-f(x)}{\Delta x}$

And I'm also stuck at this problem:

4.
Given $y=f(x)=2x^2-x$
Find $\Delta y=f(x+\Delta x)-f(x)$

Help is greatly appreciated. :)

Thanks guys!

--RS
• February 14th 2008, 08:21 AM
Greengoblin
1.) can be written as f(x) = x^(-2) - x^(1/3),
so the derivative will be:

f '(x) = -2x^(-3) - (1/3)x^(-2/3)

2.) you can you the quotient rule, where dy/dx = [V(du/dx) - U(dv/dx)] / V^2 and u = x^2, v = (1-3x)
• February 14th 2008, 08:28 AM
Quote:

Originally Posted by Greengoblin
1.) can be written as f(x) = x^(-2) - x^(1/3),
so the derivative will be:

f '(x) = -2x^(-3) - (1/3)x^(-2/3)

2.) you can you the quotient rule, where dy/dx = [V(du/dx) - U(dv/dx)] / V^2 and u = x^2, v = (1-3x)

If it is okay by you, could you explain to me how exactly did you do that? I'm sorry, derivatives are one (Swear) of a math topic I'm stuck with and my professor is those types that knows his stuff but can't really teach it to us.
• February 14th 2008, 08:53 AM
Greengoblin
Sure. First of all, the derivative of a sum of functions is the sum of derivatives of those functions, so:

y = f(x) + g(x) + h(x)
dy/dx = f '(x) + g '(x) + h '(x)

The power rule tells us that for a function like x^n:

dy/dx = nx^(n-1)

In other words we bring the power down in front of x, and then decrease the power by one.

when you have the inverse of a function x^n you have 1/ x^n, and this can also be written as x^-n. For the first term I did it like this, and then used the power rule, to get -nx^(-n-1).

For the second term you can wite square roots as fractions. For example sqrt(x) can be written x^(1/2), the cubed root can be written x^(1/3), and the nth root can be written x^(1/n). Once you have written it like this you can simply apply the power rule to it again. so than if f(x)= nth root of x:

f(x) = x^(1/n)
dy/dx = (1/n)x^(1/n - 1)

When you have a problem like this, it's much easier to write inverses out as negative powers, and square roots out as fractional powers first. Knowing this, you could probably make a good attempt at the second derivative? If not I'll try it for you. Cheers!
• February 14th 2008, 05:39 PM
mr fantastic
Quote:

Given $f(x) = \frac{x^2}{1-3x}$
The quotient rule says that if y = u(x) v(x), then $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.
In your problem, $u = x^2 \Rightarrow \frac{du}{dx} = 2x \,$ and $v = 1 - 3x \Rightarrow \frac{dv}{dx} = -3$.