# Well then ....

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Posted by Subhotosh Khan on November 11, 2002 at 17:44:32:

In Reply to: Re: What did you get for Dx[(2x-5)^4] and Dx[x^3] ?(n/t) posted by tim on November 11, 2002 at 17:19:08:

: : :
: : : : you must use a combination of rules.

: : : : start with the product rule (since you are multiplying things), then use power rule.

: : : : note that you have x^3(2x-5)^4

: : : : so you have a product of x3 and (2x-5)4

: : : : so Dx[x^3(2x-5)^4]

: : : : = x^3*Dx[(2x-5)^4] + (2x-5)^4*Dx[x^3]

: : : : where Dx[y] means the derivative of y with respect to x.

: : : I can get that far. then im stuck. when i check it on the calculator i get x²(2x-5)^3(14x-15).
: : : and i can't get that answer.

: for Dx[(2x-5)^4] i get 4(2x-5)^3*2 = 8(2x-5)
: for Dx[x^3] i get 3x²
**************************

Dx[(2x-5)^4] = 4*2*(2x-5)^3=8*(2x-5)^3

Dx[x^3(2x-5)^4] = x^3*Dx[(2x-5)^4] + (2x-5)^4*Dx[x^3]

=x^3 * 8*(2x-5)^3 + (2x-5)^4 * 3 * x^2

Now factor out x^2 *(2x-5)^3

= x^2 *(2x-5)^3 * [8*x + (2x-5)*3]

= x^2 *(2x-5)^3 * [8*x + 6x - 15]

=x^2 *(2x-5)^3 * [14x - 15]

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