One of the hardest prof (1 Viewer)

[chrisk]

given that

Θ = arctan 3/x - arctan 2/x

prove

dΘ/dx = (2 / x^2+4) - ( 3 / x^2+9)

hence find x when
dΘ/ dx = 0

incase you need an answer

ANS. x = root 6

All i need is prof of that thing... soo hardd -_-

DO NOT GIVE ME ANY LINKS..I NEED workingss.

thanks

australian international school hong kong

 

[lolokay]

i was just writing out the answer and thread got deleted

d(arctan 3/x)/dx
= (-3/x²)/(9/x² + 1)
just using derivative of inverse tan

..and you can do the rest

 

[jet]

Ok for arctan(3/x) d/dx= (1/((3/x)^2 + 1)) * d/dx(3/x)
= (1/((9/x^2) + 1)) * (-3/x^2)
=-3/(x^2 + 9)

For arctan(2/x) d/dx = (1/((2/x)^2 + 1)) * d/dx(2/x)
=(1/((4/x^2) + 1))* (-2/x^2)
=-2/(x^2 + 4)

Hence d/dx(arctan(3/x) - arctan(2/x))
= 2/(x^2 + 4) - (3/(x^2 + 9))

When d/dx = 0
2/(x^2 + 4) - (3/(x^2 + 9)) = 0

When multiplying up the denominators,
2(x^2 + 9) - 3(x^2 + 4)=0
2x^2 + 18 -3x^2 -12=0
-x^2 + 6=0
x^2 = 6
x =±root(6)

 

[chrisk]

Thanks everyone
i got it !!!! yay