Question

how do i Differentiate with respect to x
(x^2+3)^3 (2-5x)^4?

Answer 1

Do you know the chain rule?

Answer 2

kind of.. ut do i have to defferentiate the brackets separately?

Answer 3

... and the rule for products?

Answer 4

no.. !

Answer 5

What do you have to work with?

Answer 6

Im doing it with the dy/dx
=3(x2+3)^2(2x) 4(2-5x)^3(-5)

Answer 7

i dont know if that dy/dx but im finding it through the power(u)power-1 ?

Answer 8

That looks about right. Difficult to be sure. Is there a plus ahead of that 4?

Answer 9

You are using the chain rule without knowing it! ;)

Answer 10

hahahahhaa.. sorry. im so blanked right now that i cant think straight.. i need help so that y im here.. but im dong it right? and no there is no + before 4

Answer 11

So what was ahead of the 4 in the original question?

Answer 12

Don't worry. I realised the other day that it's been forty years since I did some of these things. I old and retired now.

Answer 13

the u?

Answer 14

the u in the formula? (2-5x)^4.. 4 is the exponent

Answer 15

Bill can you check my answer if it right?
-60(x^2+3)^2 (2x) (2-5x)

Answer 16

Before I do that, is this what you started with? It makes a big difference.

\[(x^2+3)^3 (2-5x)^4\]

Answer 17

The result you are offering would be more like what you'd get for this.

\[(x^2+3)^3 + (2-5x)^4\]

Answer 18

yes :D

Answer 19

oh sorry no... theres no plus in between

Answer 20

No worries but to do these you need to use the product rule.

Answer 21

which one is that..and please dont get annoyed with me not knowing.. i might know it ut dont know it call the product rule.

Answer 22

Duv = uv' + u'v

which means the derivative of a product is the sum of the first item in the product times the derivative of the second item and the derivative of the second item in the product times the first item.

It sounds much worse than it is.

I am not in the least annoyed.

Answer 23

For example,

d/dx x(1-x) = x d/dx (1-x) + (1-x) d/dx x = x (-1) + (1-x)(1) = -x + (1-x) = 1-2x.

We just need to do this for the more complicated expression you have. Takes patience.

Answer 24

i havent been taught by this actually.. i guess i'll have to do more reserch on this.. and yes you explain it wea i can understand but i have to apply thi s to my equation..

Answer 25

Im actually confuse right now.. but it my fault and i appreciate your help.. take care :D

Answer 26

There are videos on youtube that might help. Here's one: http://www.youtube.com/watch?v=uPCjqfT0Ixg. It's a lot to swallow in this format that we're using here. Come back if there's anything we can do. Sorry I couldn't help more.

Answer 27

Oh! And if you want I can just apply it to your equation and post it here.

Answer 28

okay ill look this up.. and no you were more than help.. thank you

Answer 29

oh that will be great.. but only if you have the time though

Answer 30

Just give me a few minutes.

Answer 31

sure.. thank you very much .. and also for taking your time :D

Answer 32

I hope this makes sense.

Answer 33

Wow.. it does.. Thank you :D very very much