Differentiate 2 arctan 2x?
Recall that the derivative of arctan(x) is 1/(1+x^2). Using the above formula and the chain rule, you get (2 arctan (2x))' = 2 * (1 / (1 + (2x)^2)) * 2 = 4 / (1 +4x^2)
Integral of x arctan 2x?
Using integration by parts: integral of x arctan 2x dx = integral of (1/2) arctan 2x dx^2 = (1/2)x^2 arctan 2x - (1/2) integral of 2x^2/[1+4x^2] dx = (1/2)x^2 arctan 2x - (1/4) integral of (1+4x^2 - 1)/[1+4x^2] dx = (1/2)x^2 arctan 2x - (1/4)x + (1/8)arctan 2x + c
What is the derivative of arctan (2x/1+X2)?
I assume you mean the derivative of [math]arctan(2x/(1+x^2)) [/math]The derivative is [math](2-2x^2)/(x^4+6x^2+1)[/math]
What is the nth derivative of arctan {2x/ (1-x^2)} in terms of r and theta ?
Hey guys i finally got the answer... a big thanks to david without whose insight would probably not have got it... since i dont know how to type the answer i am uploading the snaps of my notebook answer... pls excuse the poor handwriting...The sum uses eulers formula a lot... so if you are unaware of it then check it out on the net... there is ample of information available there... thanks a lot guys... math rockzzzz....
Advanced Calculus and Sigma notation?
Please show all work (pictures preferred, neat writing) 1) Find the number n such that [(n) on top of (BIG E) under e is (i=1)] of i = 78 2) Find the limit lim n->infinity of [(n) on top of (BIG E) under e is (i=1)] of 3/n [((1+(3i/n))^3)-2(1+(3i/n))] 3) Determine a region whose area is equal to the given limit. Do not evaluate the limit. lim n->infinity of [(n) on top of (BIG E) under e is (i=1)] of 2/n [(5+(2i/n))^10] 4) lim n->infinity of [(n) on top of (BIG E) under e is (i=1)] of [xi/((xi^2)+4)][delta x] and the interval is [1,3] 5) evaluate the integral by interpreting it in terms of areas integral from 0 to 10 of the absolute value of (x-5) dx 6) evaluate the integral integral from 0 to 2 of [(y-1)(2y+1)]dy 7) evaluate the integral integral from 1/2 to 1/(sqrt 2) of [4/(sqrt (1-x^2))]dx 8) find the derivative of the function F(x)= integral from sqrt x to 2x of [arctan(t)]dt 9) evaluate the integral integral from 0 to pi/3 of [(sinx+sinx((tanx)^2))/((secx)^2)]dx 10) evaluate the integral integral from 0 to 3pi/2 of the absolute value of (sinx) dx 11) evaluate the indefinite integral integral of [(arctanx)/(1+(x^2))]dx 12) evaluate the indefinite integral integral of [(sin(lnx))/x]dx 13) evaluate the indefinite integral integral of [(2^t)/((2^t)+3)]dt 14) evaluate the indefinite integral integral of [sint((sec(cost))^2)]dt
Calculus help in simplifying?
Yes, you need to factor out the smallest power which in this case is x^-1/17 and the coefficient 1/17: 1/17 * x^-1/17 [16 - 17 x^34/17] = 1/17 * x^-1/17 [ 16 - 17 x^2] This is different from what you say the answer is, so I am guessing you integrated or differentiated incorrectly. I am sure that the answer I have is the factored form of your work. You can check, by distributing back through.
How do you find [math]x[/math] in [math]\arctan(\frac x 2) +\arctan(\frac {2x} 3) =\frac {\pi} 4[/math]?
Use the arctangent addition formula:This can be proven using the tangent addition formulaNow we have arctan(x/2)+arctan(2x/3)=arctan(7x/(6–2x²))=pi/4, so 7x/(6–2x²)=tan(pi/4)=1.7x=6–2x² is a quadratic equation that is easy to solve; it has the two solutionsThe first one does not work: arctan(x/2)+arctan(2x/3)=-3/4*pi which is not equal to pi/4. (It is equal to pi/4 modulo pi, though).But the second solution actually works. So, here’s your answer: