finding derivative

• October 9th 2007, 02:59 PM
rmn
finding derivative
y= ((2x-5)^-1 ) * (x^(2)-5x)^6

this is what i have so far

dy/dx=

(2x-5)^-1 * 6(x^2-5x)^5 * (2x-5) * 2 +

(x^2-5x)^6 * -1(2x-5)^-2 * 2

cancel the (2x-5) in the first term right

so then i have

12(x^2-5x)^5- 2(x^2-5x)^6 * (2x-5)^-2

factor out 2(x^2-5x)^5

so i have

2(x^2-5x)^5 [ 6-(x^2-5x)*(2-5)^-2]

now i am confused how to simplify this
• October 9th 2007, 03:02 PM
Jhevon
Quote:

Originally Posted by rmn
y= ((2x-5)^-1 ) * (x^(2)-5x)^6

this is what i have so far

dy/dx=

(2x-5)^-1 * 6(x^2-5x)^5 * (2x-5) * 2 +

(x^2-5x)^6 * -1(2x-5)^-2 * 2

what is that 2 for?
• October 9th 2007, 03:11 PM
rmn
Quote:

Originally Posted by Jhevon
what is that 2 for?

derivative of the 2x-5
yeah i wasn't sure if i was supposed to do that?
is it part of the chain rule or something
• October 9th 2007, 03:14 PM
Jhevon
Quote:

Originally Posted by rmn
derivative of the 2x-5
yeah i wasn't sure if i was supposed to do that?
is it part of the chain rule or something

no. the chain rule was done when you wrote (2x - 5). you don't need to take the derivative of that.

so you should have: $(2x-5)^{-1}(6) \left( x^2-5x \right)^5(2x-5) + (x^2-5x)^6 (-1)(2x-5)^{-2}(2)$

now continue. what do we have common here? what can we factor out? and what powers should we factor out?
• October 9th 2007, 03:16 PM
rmn
Quote:

Originally Posted by Jhevon
no. the chain rule was done when you wrote (2x - 5). you don't need to take the derivative of that.

so you should have: $(2x-5)^{-1}(6) \left( x^2-5x \right)^5(2x-5) + (x^2-5x)^6 (-1)(2x-5)^{-2}(2)$

now continue. what do we have common here? what can we factor out? and what powers should we factor out?

oh, just a general question how do we know when to stop using the chain rule like in that question
• October 9th 2007, 03:18 PM
Jhevon
Quote:

Originally Posted by rmn
oh, just a general question how do we know when to stop using the chain rule like in that question

the chain rule says: $\frac d{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$ ........we take the derivative of the inside function once. we don't take the derivative of the derivative of g(x) (that would be g''(x), and that's not in the formula). (the only time we may have to do that is if g(x) itself is a composite function, which was not the case here)
• October 9th 2007, 03:22 PM
rmn
Quote:

Originally Posted by Jhevon
the chain rule says: $\frac d{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$ ........we take the derivative of the inside function once. we don't take the derivative of the derivative of g(x) (that would be g''(x), and that's not in the formula). (the only time we may have to do that is if g(x) itself is a composite function, which was not the case here)

oh ok thanks
ok so now i have

2(x^2-5x)^5 [ 3- (x^2-5x)(2x-5)^-2]

is that right
• October 9th 2007, 03:26 PM
Jhevon
Quote:

Originally Posted by rmn
oh ok thanks
ok so now i have

2(x^2-5x)^5 [ 3- (x^2-5x)(2x-5)^-2]

is that right

yes, that's fine