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davidabrown
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Problem Set 2, Integrations techniques, Problem 5A3g. Can someone please explain this in layman's terms?
 one year ago
 one year ago
davidabrown Group Title
Problem Set 2, Integrations techniques, Problem 5A3g. Can someone please explain this in layman's terms?
 one year ago
 one year ago

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davidabrown Group TitleBest ResponseYou've already chosen the best response.0
Problem Set 2, Intgegrations techniques, Problem 5A3h. I am having the same problem. I can't seem to understand what the solution is trying to explain. Please advise.
 one year ago

jsghetti Group TitleBest ResponseYou've already chosen the best response.0
The first line is stating known facts about the function inside of the inverse tangent (3g), using y as a shorthand for the function to make the math a little more intuitive. When we rewrite it as \[\tan^{1} y\]it becomes much more intuitive to write the derivative of the function as \[\ \frac{ \frac{ dy }{ dx } }{ 1+y^2 }\]. We have to have the dy/dx in the top because y is still a function of x, and we are taking the derivative with respect to x, so that function has to have the derivative taken according to the chain rule, and while we could substitute in y in the bottom, it makes it much easier to manipulate our original y to fit that denominator in this case, hence\[y=\frac{ x }{ \sqrt{1x^2}}\]\[y^2=\frac{ x^2 }{ 1x^2 }\]\[1+y^2=\frac{ x^2 }{ 1x^2 }+1\]which we then have to get a similar denominator to add these, which will finally yield the stated equation on the first line of the solution\[1+y^2=\frac{ 1 }{ 1x^2 }\]From there, we just substitute everything in to the derivative, to get the stated intermediate step, and simplify. Hope that helps for 3g, let me know if not and I will try to write it out better for you. 3h uses a similar technique, I could write it out for you if you need still, but I think this solution and that one are extremely similar.
 one year ago
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