# Differentiating a fraction

• Jan 23rd 2008, 11:40 AM
Smithswood
Differentiating a fraction
ok i have this question and need some urgent help, i have to hand this in tomorrow and im rapidly falling very tired;

i am studying AS level mathematics on the aqa syllabus...

i have this question for homework:

A curve has the equation: $\displaystyle Y= X^4/4 + 32/X^2$

A: find $\displaystyle dy/dx$

i dont simply want the answer is it possible for anyone to give me a worked solution?

• Jan 23rd 2008, 11:53 AM
mr fantastic
Quote:

Originally Posted by Smithswood
ok i have this question and need some urgent help, i have to hand this in tomorrow and im rapidly falling very tired;

i am studying AS level mathematics on the aqa syllabus...

i have this question for homework:

A curve has the equation: Y= X^4/4 + 32/X^2

A: find dy/dx

i dont simply want the answer is it possible for anyone to give me a worked solution?

$\displaystyle y = \frac{x^4}{4} + \frac{32}{x^2} = \frac{x^4}{4} + 32 x^{-2}$.

Now apply the well known rule (to each term) that the derivative of $\displaystyle a x^n$ is $\displaystyle n a x^{n-1}$.
• Jan 23rd 2008, 12:04 PM
Smithswood
so...?
so$\displaystyle \frac{x^4}{4}$

would become $\displaystyle 4x^3$

and how did you get $\displaystyle 32x^-2$?
• Jan 23rd 2008, 12:12 PM
a tutor
Quote:

Originally Posted by Smithswood
so$\displaystyle \frac{x^4}{4}$
would become $\displaystyle 4x^3$

No it would become $\displaystyle \frac{4x^3}{4}=x^3$

Quote:

Originally Posted by Smithswood
and how did you get $\displaystyle 32x^-2$?

$\displaystyle \frac{a}{b^n}=ab^{-n}$
• Jan 23rd 2008, 12:44 PM
Smithswood
therefore...
would it not be $\displaystyle 32x^-1$

as $\displaystyle 32 = 32x^1$?

$\displaystyle 1 - 2 = -1$???
• Jan 23rd 2008, 12:57 PM
a tutor
Quote:

Originally Posted by Smithswood
would it not be $\displaystyle 32x^-1$

No $\displaystyle \frac{32}{x^2}=32x^{-2}$ Note, this has not been differentiated yet.