Question

I am having trouble grasping the concept of the chain rule and using it for substitution when integrating

Answer 1

well how well do you understand the chain rule for derivatives?

Answer 2

maybe you should just post a specific question...

Answer 3

I understand it for equations such as\[2x(x ^{2}+1)^{4} and \sqrt{2x+1}\]. I do not understand problems such as \[1\div \sqrt{1-9x ^{2}}\] or\[\left( y ^{2} \right)\div \left( y+1 \right)^{4}\]. I get the easy problems. I have been trying and trying to redo math problems for a test tomorrow.

Answer 4

Yikes, gotta get the chain rule for derivatives down first...
I gotta go
I'll have to let SmoothMath help you out.

Answer 5

derivative of ln(x) is 1/x. Any integral that has 1/(SOMETHING) is just begging for a ln(something) antiderivative.

Answer 6

alright. Let me work this out and see what I come up with. hold on.

Answer 7

Actually, it turns out in this case ln is not the answer.

Answer 8

So in this case.. what do you do?

Answer 9

Okay, so a lot of these REALLY SCARY ones...
have some magical thing that you just have to find that makes it work out.

Answer 10

It helps to have a list of derivative rules in front of you for that kind of problem. Then you can look down the derivatives column and find the one that looks like the integral you're doing at the moment. In this case, hopefully you'll notice that the derivative for arcsin(x) is 1/sqrt(1-x^2)

Answer 11

Problem is, I am not allowed to have the list. I know that it is \[\sin^{-1} \]\[\left( \sin^{-1} 3x \right)/3\]. That is not the answer I came up with. I did not know to use \[\sin^{-1} \]

Answer 12

Well then you'll have to memorize all of those ugly derivative =( but here's the process if you know them.

Answer 13

\[1 \div \sqrt{1-9x^2}\]
"Oh, hey! That looks kind of like the same form for the derivative of arcsin, which I know is:"
\[1\div \sqrt{1-x^2}\]
So maybe the antiderivative is arcsin(x)... Well. Not x, because that wouldn't give 1-9x^2. What could I choose instead of x so that arcsin' was 1/sqrt(1-9x^2)?

Answer 14

Okay, hey! I could choose arcsin(3x) instead. According to the derivative formula for arcsin, that gives:
\[1 \div \sqrt{1-(3x)^2} = 1 \div \sqrt{1-9x^2}\]
but wait! Chain rule says that I have to then multiply that by the derivative of the inside function, which was 3x. So the derivative now becomes:
\[3*(1 \div \sqrt{1-9x^2})\]
How can I change the function to get rid of that 3? I guess I'll add a coefficient of 1/3 so that it will cancel when I take the derivative. So now I have:
\[(1/3) \arcsin(3x)\]

Answer 15

Do you understand what I did and why? Ask me questions.