The Derivative of (1-x^2)/(x^3)?

Question

How do we differentiaite m(x) = (1-x^2)/x^3 ? please show steps.

eveyone says something different i'm confused...

Answer 1

Okay so you have to break a part the fraction and then take the derivative of both.

d/dx 1-x^2 times d/dx 1/x^3

now the first derivative you can break it apart into the derivative of 1 - the derivative of x^2 which gives you

f'1 = 0 f' (x^2)= 2x so that's 0-2x which is -2x

the second derivative is d/dx of X^-3 (which is the same as 1/x^3) and that would be f' = -3x^-4 so your answer is both multiplied.

-2x times -3x^-4 and then you can multiply.  

Answer 2

Ignore the responses below...

From what I've seen only one person, James, is actually correct.   If people are uncertain its better not to post because it will only serve to confuse a person learning.  

Instead let me show you how this done step by step so that you may learn from it.  

First of all its a good idea to write down all of the derivative rules you are currently learning.  My guess is that this homework problem revolves around one of them.  If you look at them you will see one that immediatelly sticks out...  We have a fraction or Quotient...

Thus we nee "The Quotient Rule"

Here is a link to it...

http://en.wikipedia.org/wiki/Quotient_ru...

It states...

For  

f(x)  = g(x) / h(x)

f ' (x) =

g' (x) h(x) - h'(x) g(x)

-------------------------------

( h(x) ) ²

Thus...

Let

g(x) = 1-x²

g'(x) = - 2x

Let

h(x) = x^3

h'(x) = 3x²

Now substitute for our general equation

g' (x) h(x) - h'(x) g(x)

-------------------------------

( h(x) ) ²

Which gives us...

(-2x)(x^3) - (1 - x²)(3x²)

----------------------------------

(x^3)²

Simplify by distributing

-x^4 - (3x² - 3x^4)

----------------------------

x^6

Simplify by distributing the negative

-x^4 - 3x² + 3x^4

----------------------------

x^6

Collect like terms

2x^4 - 3x²

----------------------------

x^6

Factor out x² from both top and bottom

x²( 2x² - 3)

-----------

x^4 (x²)

Cancel terms and we are done..

2x² - 3

-----------

x^4

This is the answer with 100% certainty.

For future reference I suggest looking searching on the internet for derivative rules.. there are lots of good sites with links to general equations for both derivatives and integrals.

I use them on a daily basis as I am a physics programmer...

Answer 3

y = x^(- 3) - x^(- 1)

dy/dx = - 3 x^(- 4) + x^(- 2)

dy/dx = - 3 / x^4 + 1 / x²

Answer 4

For this Calculus problem involving derivatives, you do not need to listen to what people say.  Instead listen to the truth, which is the rules of Calculus.  Calculus is math and think math when approaching such problem not human.  Now in commencing with this problem, you have an algebraic expression of the form 1-x^2 / x^3, which is equivalent to a rational function of N(x) / D(x) = f(x).  When differentiating a rational function look for connections; that is, how does this rational function relate to the rules of derivatives.  If you are given a rational function f(x) = g(x) / h(x), the derivative of the function itself involving quotients or rationals is: d/dx[g(x)/h(x)] = d/dx[f(x)] = h(x)g(x)' - g(x)h(x)' / h(x)^2.  Thus, the derivative of this function fits the rule of differentiating quotients because f(x) = g / h; where g = 1 - x^2 && h = x^3.  Thus, the derivative of f is: x^3(-2x) - (1 - x^2)(3x^2) / x^6.  This simplifies to -2x^4 -3x^2 + 3x^4 / x^6 = x^4 - 3x^2 / x^6.  Now you can factor the numerator to f'(x) = x^2(x^2 - 3) / x^6 = x^2 - 3 / x^4.  Thus, the derivative of f(x) = 1 - x^2 / x^3 = x^2 - 3 / x^4.  To know if this is true use the fundamental theorem of calculus by doing the inverse operation of differentiation, which is integration.

J.C

Answer 5

let f(x)=(1-x²) and g(x)=x³

D(f(x)/g(x))= (f'(x)g(x)-g'(x)f(x))/(g(x))²

which means you differentiate the numerator, 1-x² to get f'(x)=0-2x=-2x and you differentiate the denominator to get g'(x)=3x²

Now simply apply the above rule

D(f(x)/g(x))=D((1-x²)/x³)) =(f'(x)g(x)-g'(x)f(x))/(g(x))²

=(-2x(x³)-3x²(1-x²))/(x³)²

You can simplify if you want to.