Hello Again Steve I'm sorry to trouble you I wanted to ask 2 questions again if its okay with you. I promise to leave a higher bonus this time.1.) Differentiate y=x(arctan(2x))^2 2.) Find ∫2dx/√9-4x^2I don't need it as immediately as last time but if you can give it to me sometime in the afternoon sir I will greatly appreciate it.
hello there! can I help you? I am now starting working on these dear customer, thank you for using justanswer.com
1.) y = x[arctan (2x)]^2using chain rule (with product rule)dy/dx = 1* [arctan (2x)]^2 + x*2*[arctan(2x)][2 /(1 + 4x^2)]dy/dx = [arctan (2x)]^2 + 4x[arctan(2x)]/(1 + 4x^2)2.)∫2dx/√9-4x^2 = arcsin(2x/3) + CPlease Leave a POSITIVE FEEDBACK to Credit me. BONUS is welcome.Thanks.If you need my help asap, just write in the first part of your question: "For Mr Glenn G. only!" Thanks againPlease don't forget to give me a positive rating, thanks a lot!
To the expert:
Is not allowed to reply to questions with some name , in this case "For Steve"
To the customer:
I will post the answer today, please ignore the answer above, is not allowed for an exper to do that since you requested me "For Steve"
To Dear esteemed expert Steve, Since you were requested, there is a 15 mins lock in period for you that no can answer or reply the customer, but it says that after that 15 min lock in period expires, that is the time any available expert online can help the customer and post the answer, since the customer also needs to be attended, and the privilege of the 15 minute grace period/waiting time given to the requested expert has already been expired, also, there is no rule about that if a given expert is requested, and happens to be not available, then still that posted question with a requested expert will be only /exclusively and absolutely be given to that specific requested expert, there is no rules about that, just answer only give the requested expert at most 15mins to reply or else, other available experts can replace him/her and attend to the customer. Hope for your understanding and consideration. (I have waited patiently for that 15-minute lock in period to expire just in case you were not online my dear co expert) So please do not disregard my posted answer, please be fair also since there are a lot of other experts who are patiently online and waiting to help customers, just in case the requested expert is not available. The rules of justanswer.com I think is clear, I hope so. Respectfully yours, TrigeoscalMr. Glenn G.41085.6307677893
Ok Sir Is there any other steps(work) required for the question or is that it? I need the steps for refrence only.
for question 1:using product rule of differentiation:if y = u*v, then dy = u*dv + du*vso the derivative of the product (of two functions) is equal to the derivative of the first factor* the second factor plus the second factor*the derivative of the second factor.for example: y = x^2*sinx, so the derivative of y with respect to x is = d(x^2) * sinx + x^2 * d(sinx)= 2xsinx + x^2cosxthen in differentiating the second factor in the given, it is (arctan 2x) we will use the chain rule for differentiation;that is y = f(g(x), then y' = g'(x)*f'(g(x)so the derivative of z = [arctan 2x]^2,let g(x) = arctan 2x and f(x) = x^2, theng'(x) = 2/(1 + 4x^2) and f'(x) = 2x, and soz = [arctan 2x]^2 = g'(x)*f'(g(x)) = 2*arctan2x * 2/(1 + 4x^2) = (4xarctan 2x) / (1 + 4x^2)for number 2:the formula for ∫adx/√(b^2-e^2x^2) where a, b and e are constants is ∫adx/√(b^2-e^2x^2) = (a/e)*arcsin(ex/b) + Cexample:∫3dx/√(25- 16x^2) = (3/4)*arcsin(4x/5) + Cand ∫5dx/√(49- 36x^2) = (5/6)*arcsin(6x/7) + CPlease don't forget to press accept if you are satisfied, so that I will be compensated for my time and effort in helping youPlease Leave a POSITIVE FEEDBACK to Credit me. BONUS is welcome.Thanks.If you need my help asap, just write in the first part of your question: "For Mr Glenn G. only!" Thanks again
Again : Is not allowed to reply (is my name on the question!!!!)
Here are my answers
I appreciate if you accept/rate my answers (since you requested me)
I teach Calculus and Probability in an University since 1994
(The note above was for the expert)