h(x)=2 squarex + 3 3 square x

k(t)= 8t 3/2

g(x) 6x^5-4x^3+x^2

Printable View

- Mar 25th 2006, 01:45 PMbatman123derivative
h(x)=2 squarex + 3 3 square x

k(t)= 8t 3/2

g(x) 6x^5-4x^3+x^2 - Mar 25th 2006, 01:48 PMTD!
And the question is, find the derivative?

Do you know the different rules and if so, what is the problem? - Mar 25th 2006, 04:02 PMThePerfectHackerQuote:

Originally Posted by**batman123**

Given,

$\displaystyle 6x^5-4x^3+x^2$

You need to find, (the derivative)

$\displaystyle (6x^5-4x^3+x^2)'$

thus, by the sum-rule,

$\displaystyle (6x^5)'-(4x^3)'+(x^2)'$

thus, by the constant-multiple rule,

$\displaystyle 6(x^5)'-4(x^3)'+(x^2)'$

thus, by the power-rule,

$\displaystyle 6(5x^4)-4(3x^2)+(2x)$

thus, after some algebra,

$\displaystyle 30x^4-12x^2+2x$

$\displaystyle \left<\begin{array}{ccc}\mathbb{Q}&\,&\, \\ \,& \mathbb{E}&\, \\ \,&\,&\mathbb{D} \end{array}\right>$