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anonymous
 one year ago
Can someone show me how to find
d/dx (arctan (5x))
anonymous
 one year ago
Can someone show me how to find d/dx (arctan (5x))

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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{d}{dx} \tan^{1}(5x) \] this?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1ok. the formula for the derivative of the inverse trigonometric function is \[\tan^{1}(x) = \frac{1}{1+x^2}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1so we need to find the derivative of 5x first what is the derivative of 5x? @cstarxq

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1I'm not using the right variable here...this is going to be confusing a bit \[\frac{d}{dy} \]\[\tan^{1}(y) = \frac{1}{1+y^2}\] let y = 5x so we need the derivative of 5x which is?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1mhm that's correct so that goes in the front and then afterwards you just substitute y=5x in the fraction

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{5}{1+y^2}\] so if y =5x then we have (5x)^2 which is

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1(5x)^2 = (5x)(5x) = ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1correct and that's it.. \[\frac{5}{1+25x^2}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1you just have to take the derivative of the inside for \[\frac{d}{dx} \tan^{1}(5x) \] in that case it was a 5x so the derivative of that is 5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh i see, i forgot i had to take the derivative of 5x and substituting that into y^2..

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1then the derivative goes on the numerator.. and then the denominator is just pure substitution

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1can I get a medal ? ^^

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks for answering my question!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1you're welcome :) If I remembered correctly, you need to memorize them >.< Calculus II had so much formulas.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x____x oh, yeah it does

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1Calculus III should be easy, but Calculus IV... UGH! So many conversations, integrations, and vectors come back

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I lost my connection, but I'm taking calc III and linear algebra next semester. Hopefully it won't be too bad.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1I've already taken those subjects. It's not too bad

idku
 one year ago
Best ResponseYou've already chosen the best response.0whether you are still viewing or not, there is a way of doing it. Don't be a slave of a nonmaking sense (or even making sense) formula. Every formula you use should be something you understand and that is what I will try to improve right now.

idku
 one year ago
Best ResponseYou've already chosen the best response.0you are dealing with an inverse function. Let's say inv. tan. \(y={\rm Tan}^{1}(x)\) rewrite it first. \(x={\rm Tan}^{}(y)\) (i hope you know why I did this, or elsego back and rvw that rule) Now differentiate with respect to x. (Taking a derivative of a function y=g(x), that is dy/dx) \(\frac{dy}{dx}~x=\frac{dy}{dx}~{\rm Tan}^{}(y)\) \(1={\rm Sec}^{2}(y)\times y'\) \(1/{\rm Sec}^{2}(y)= y'\) \(1/\left({\rm Tan}^{2}(y)+1\right) = y'\) you know what tan(y) is from the beginning, and thus you know what tan^2(y) is as well. \(1/\left(x^2+1\right) = y'\) and there is your derivative

idku
 one year ago
Best ResponseYou've already chosen the best response.0same you can do with any inverse function.....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I'm still reviewing this. Thanks @idku, I just used your method to evaluate the derivative for the original problem I was trying to figure out which was actually using integration by parts to evaluate integral of arctan(5x)...
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