If y=tanx then prove that d^2y/dx^2=2y(1+y^2)?
Answer:
Proving this can get really ugly and since I'm not sure how far you are in math, I'm going to show all the steps (so sorry if some of these steps bore you).
First, before we attempt to prove your question, we should set up a useful equality. We know that:
(sinx)^2 + (cosx)^2 = 1
By manipulating the above equation, we get the following steps:
1) (sinx)^2 = 1 - (cosx)^2
[subtract (cosx)^2 from both sides]
2) (sinx)^2 / (cosx)^2 = [1 - (cosx)^2] / (cosx)^2
[divide both sides by (cosx)^2]
3) (tanx)^2 = [1/(cosx)^2] - 1
[On the left hand side, we use the equality that tanx = sinx/cosx, and on the right hand side, we simply divide both the "1" and "(cosx)^2" by "(cosx)^2]
4) (tanx)^2 +1 = 1/(cosx)^2
[add 1 to both sides]
5) (tanx)^2 +1 = (secx)^2
[use the identity: secx = 1/cosx. Also: sec = secant.]
Thus, from (sinx)^2 + (cosx)^2 = 1, we can show that (secx)^2 = (1 + (tanx)^2)
Now that that is out of the way, we can start proving that if y = tanx, d^2y/dx^2 = 2tanx(1+(tanx)^2).
So:
Let y = tanx
1) y = sinx / cosx
[using the identity that tanx = sinx/cosx]
2) y' = dy/dx = [cosx*cosx - sinx*(-sinx)] / (cosx)^2
[Use the quotient rule. Also note that "*" is the same as the multiplication sign]
3) y' = [(cosx)^2 + (sinx)^2] / (cosx)^2
[simplify the equation]
4) y' = 1 / (cosx)^2
[Since (cosx)^2 + (sinx)^2 = 1, we know that the numerator is equal to 1.]
5) y' = dy/dx = (secx)^2
This is our first derivative for y = tanx.
6) y' = 1 / (cosx)^2
[using the equality that cosx = 1/secx.]
7) y'' = d^2y/dx^2 = [(cosx)^2 * (0) - (1)*2(cosx)*(-sinx)] / (cosx)^4
[Use the quotient rule]
8) y'' = [2(cosx)(sinx)] / (cosx)^4
[Since (cosx)^2*(0) =0, we can simply remove that term from the equation to simplify it.]
9) y'' = 2 sinx / (cosx)^3
[Simply divide the numerator by the denominator, ONE of the cosx's cancels out (leaving 3 cosx's on the denominator)]
10) y'' = [2 (sinx/cosx)] / (cosx)^2
[Part 10 and Part 9 are completely identical. I merely grouped it differently so Part 11 is more easy to understand. If the notation confuses you, just skip part 10]
11) y'' = 2tanx / (cosx)^2
[using the identity sinx/cosx = tanx]
12) y'' = d^2y/dx^2 = 2tanx*(secx)^2
[Using the identity secx = 1/cosx]
This is our second derivative. The last step to this problem is to rewrite it so that it looks like 2y(1+y^2).
Thus:
12) y'' = d^2y/dx^2 = 2 tanx * (secx)^2
13) y'' = 2*tanx*[1+ (tanx)^2]
[Remember what we did at the beginning? We simply substituted "(secx)^2" with "1+(tanx)^2"]
14) y'' = 2y(1+y^2)
[Since we are given that y= tanx, substitute all "tanx" with "y"]
Thus, we reach the result that if y = tanx, then y'', or d^2y/dx^2 = 2y(1+y^2).
Hope this helps.
Use the triangle which having angle x and the opposite side equals to y and the adjacent side is 1. Then, by trigonometry tan x = y / 1
or simply y=tanx.
then y =sinx/cosx , take the first derivative y' = (cosx(cosx) -sinx(-sinx))/(cosx)^2
simplify it, becomes y' = 1/(cosx)^2
then again take the 2nd derivative y'' = [((cosx)^2)(0) - 1(2 cosx)(-sinx)]/(cosx)^4
simplify becomes, y'' = 2sinx / (cosx)^3
or y'' = 2 tanx / (cosx)^2
go back to the triangle,
tanx = y
cosx = 1 / (1 + y^2)^1/2
then 1/(cosx)^2 =1+y^2
substitute this value so, y'' = 2y(1+y^2)
its simple
y=tan x .......given
differentiate with respect x we get
dy/dx=(sec x)^2 .........
but (sec x)^2 = 1+(tan x)^2
hence (sec x)^2 = 1+y^2
hence substituting in equation A
dy/dx = 1+y^2 .........
differentiate with respect x we get
d^2y/dx^2 = 2y dy/dx
d^2y/dx^2 = 2y (1+y^2) ......... 1
hence the proof
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: