The Student Room Group

C3 Trig Diff

show that d/dx (ln tan 0.5x) = 1/sinx

I just want to see where im going wrong

d/dx ln f(x) = f'(x)/f(x)

so tan0.5x = sec20.5x

so d/dx ln f(x) = sec20.5x/tan 0.5x

= 1/cos20.5x/( sin 0.5x/ cos 0.5x)

multiply both sides by cos 0.5x

cos 0.5x / cos20.5x / (sin 0.5x)

cancel nominators cos 0.5x

1/cos 0.5x / sin0.5x

multiply both sides by cos 0.5x

1/(sin 0.5x)(cos0.5x)

okay so now im scratching my head thinking what next when it hits me.

sin 2x = 2sinxcosx

so 1/2 x sin2x = sinxcosx

in this case x = 0.5x

so 1/(sin 0.5x)(cos0.5x) = 1/0.5sinx = 2/sinx

So can someone check what ive done and point the error out thanks.
insparato
show that d/dx (ln tan 0.5x) = 1/sinx

I just want to see where im going wrong

d/dx ln f(x) = f'(x)/f(x)

so tan0.5x = sec20.5x

so d/dx ln f(x) = sec20.5x/tan 0.5x

= 1/cos20.5x/( sin 0.5x/ cos 0.5x)

multiply both sides by cos 0.5x

cos 0.5x / cos20.5x / (sin 0.5x)

cancel nominators cos 0.5x

1/cos 0.5x / sin0.5x

multiply both sides by cos 0.5x

1/(sin 0.5x)(cos0.5x)

okay so now im scratching my head thinking what next when it hits me.

sin 2x = 2sinxcosx

so 1/2 x sin2x = sinxcosx

in this case x = 0.5x

so 1/(sin 0.5x)(cos0.5x) = 1/0.5sinx = 2/sinx

So can someone check what ive done and point the error out thanks.


integral of tan0.5x = 0.5sec20.5x

you may kick yourself now
Reply 2
Doh!! thanks.
Reply 3
silent ninja
integral of tan0.5x = 0.5sec20.5x

you may kick yourself now


tanx is differentiated to give sec²x, not integrated. :P
Reply 4
Yeah :biggrin:. Integration of trig isnt until C4.
SunGod87
tanx is differentiated to give sec²x, not integrated. :P


oops.
Reply 6
No i dont mind, yeah silly me hehe. Anyways heres like a C3/C2 diff question i just dont get where i have three answers when its two in the back.


Find the coordinates of the maximum and minimum points on the curve
y = x2(5-x)3. distinguishing between them.

Okay if y = x2(5-x)3

chain rule ( i actually did it right :biggrin:)

dy/dx = x2x 3(5-x)2 x -1 + (5-x)3 x 2x

dy/dx = -3x2(5-x)2 + 2x(5-x)3

the coordinates of a max and min point happen when dy/dx = 0

so -3x2(5-x)2 + 2x(5-x)3 = 0

take out x(5-x)2 as a factor

x(5-x)2(-3x + 2(5-x)) = 0

x(5-x)2(-5x + 10) = 0

so x = 0,2,5

and you can sub the values into y to get the coordinates and do second derivative to find out the max and min.

In the back of the book it only allows x = 0,2

am i doing something wrong here?
x=5 looks like a point of inflexion to me.
I see. And you telling him that at x=5, dy/dx = 0, but that x=5 isn't a maximum or a minimum isn't giving the game away? :p:
Reply 9
Some how i thought it would be like that lol. I just did the second derivative :rolleyes:. Thanks means im not going insane hehe :biggrin:.

Mental note watch out for point of inflexions :biggrin:.
insparato
Some how i thought it would be like that lol. I just did the second derivative :rolleyes:. Thanks means im not going insane hehe :biggrin:.

Mental note watch out for point of inflexions :biggrin:.

Well... at points of inflexion the second derivative will be 0. So you may well be going insane. :p: