Let x = ln (1 + T ^ 2), y = t-arctant. Find (d ^ 2Y) / (DX ^ 2) In addition, if we find dy / DX ^ 2, d ^ 2Y / DX, what is the calculation?

Let x = ln (1 + T ^ 2), y = t-arctant. Find (d ^ 2Y) / (DX ^ 2) In addition, if we find dy / DX ^ 2, d ^ 2Y / DX, what is the calculation?

Dy / DX = [1-1 / (1 + T?) / [2T / (1 + T?)] = t / 2D? Y / DX? = (1 / 2) * DT / DX = (1 / 2) / (DX / DT) = (1 / 2) / [2T / (1 + T?)] = (1 + T?) / (4T) I hope it can help you. If you solve the problem, please click the "select as satisfied answer" button below

How to calculate the - 1 power of a number

reciprocal

Calculate the 2007 power of (- 0.25) times the 2008 power of 4

Hello!
The 2007 power of (- 0.25) times the 2008 power of 4
=- 0.25 to the 2007 power x 4
=-(0.25x4) to the 2007 power X4
=-1x4
=-4
If you don't understand this question, you can ask again. If you are satisfied, please click "select as satisfactory answer"
If you have any other questions, please accept this question and send it to me for help
Wish you progress!

Cos squared minus sin (- α) Tan (360 degrees + α) Simplify!

first
cosx=cos(-x)
sin(-x)= - sinx
Tan (360 + x) = TaNx. The period of TaNx is π, that is 180 degrees
So the original formula is changed into
cos²a+(1/cosa)

Calculate (√ 5-2) 2011 power * (√ 5 + 2) 2010 power

(√ 5-2) 2011 power * (√ 5 + 2) 2010 power
=(√ 5-2) * (√ 5-2) 2010 power * (√ 5 + 2) 2010 power
=(√5-2)*[(√5-2)*(√5+2)]^2010
=(√5-2)*1^2010
=(√5-2)*1
=√5-2
Hope to help you

Simplification. Sin α cos α (Tan α + cos α)

sinαcosα(tanα+cosα)
=sinαcosα(sinα/cosα+cosα)
=sin²α＋sinαcos²α

Simplification: cos (π / 2 + α) - Tan (π / 2 - α) sin (π / 2 + α)

cos(π/2+α)-tan(π/2-α)sin(π/2+α)
=-sinα-cotαcosα
=-sinα-(cosα/sinα)cosα
=-(sin²α+cos²α)/sinα
=-1/sinα
=-cscα

Simplification of COS (π - α) sin (α + 2 π) / cos ^ 2 (- α - π) Tan (2 π - α)

cos(π-α)sin(α+2π)/cos^2(-α-π)tan(2π-α)
=-cosasina/cos^2a*tan(-a)
=-sina/cosa*(-tana)
=(tana)^2

Simplify the - 1 power of [sin square (π + a) cos (π / 2-A) + Tan (2x-a) cos (- a)] / - Sin square (- a) + Tan (- π + a) [Tan (a + π)]

=[(sina)^3-sina]/-(sina)^2+1
=-sina

If the quadratic power of x = (- radical 3), then x is equal to

± radical 3