The Student Room Group
Reply 1
to find c) you add the two results for a) and halve the answer...

{secxtanx -cosecxcotx}/2
Reply 2
d [ 2e^(-3x).cos[2x] ]/dx = d[2e^(-3x)]/dx .cos[2x] + 2e^(-3x).d[cos[2x]]/dx
...................................dy/dx = -6e^(-3x).cos[2x] - 4e^(-3x).sin[2x]
...................................dy/dx = -e^(-3x)(6cos[2x] + 4sin[2x])
...................................dy/dx = -2e^(-3x)(3cos[2x] + 2sin[2x])

sub (x=0) into dy/dx = -2e^(-3(0))(3cos[2(0)] + 2sin[2(0)])
............................= -6
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y = secx = 1/cosx

Result using implicit differentiation:

y = 1/cosx
ycosx = 1
dy/dx.cosx - ysinx = 0
dy/dx.cosx = ysinx
dy/dx = ysinx/cosx, Now sub in y = 1/cosx ...
dy/dx = sinx/cos²x

dy/dx = (sinx/cosx)(1/cosx) = tanxsecx
--------------
y = cosecx = 1/sinx

Using Implicit Differentiation:

y = 1/sinx
ysinx = 1
dy/dx.sinx + ycosx = 0
dy/dx.sinx = - ycosx
dy/dx = - ycosx/sinx Now sub in y = 1/sinx ...
dy/dx = -(1/sinx)(cosx/sinx)

dy/dx = -cosec[x]cot[x]
--------------
cosecx + secx = 1/sinx + 1/cosx

1/sinx + 1/cosx = (cosx + sinx) / sin[x]cos[x]

Now use identity 2sin[x]cos[x] = sin[2x]

sin[x]cos[x] = ½sin[2x] , sub this result into equation:

.... (cosx + sinx) / ½sin[2x]
.... 2(cosx + sinx) / sin[2x]
Reply 3
fallen_star


Here is my solution for the first part. It may be wrong. It does not seem to be the same as Ricki's

Here is the graph. I have used the gradient that Ricki found i.e - 6
Reply 4
d/dx [ (sinx + cosx)/sin[2x] ] = d/dx [ ½(1/sinx + 1/cosx) ]
...................................... = ½ (-cosec[x]cot[x] + tanxsecx) (using results from previous parts)
....................................... = ½(tanxsecx - cosec[x]cot[x])
....................................... = ½(sinx/cosx.1/cosx - 1/sinx.cosx/sinx)
....................................... = ½(sinx/cos²x - cosx/sin²x)
............................... dy/dx = (sin³x - cos³x)/(2cos²[x]sin²[x])

Simplification is not necessary if not needed, for dy/dx.
Reply 5
wow the solutions seem complicated
Reply 6
lil_kk
wow the solutions seem complicated


Here I have graphed the computers result and Ricki's result. The graphs overlap ... which shows both solutions are correct.
Reply 7
Ricki
d [ 2e^(-3x).cos[2x] ]/dx = d[2e^(-3x)]/dx .cos[2x] + 2e^(-3x).d[cos[2x]]/dx
...................................dy/dx = -6e^(-3x).cos[2x] - 4e^(-3x).sin[2x]
...................................dy/dx = -e^(-3x)(6cos[2x] + 4sin[2x])
...................................dy/dx = -2e^(-3x)(3cos[2x] + 2sin[2x])

sub (x=0) into dy/dx = -2e^(-3(0))(3cos[2(0)] + 2sin[2(0)])
............................= -6
--------------
y = secx = 1/cosx

Result using implicit differentiation:

y = 1/cosx
ycosx = 1
dy/dx.cosx - ysinx = 0
dy/dx.cosx = ysinx
dy/dx = ysinx/cosx, Now sub in y = 1/cosx ...
dy/dx = sinx/cos²x

dy/dx = (sinx/cosx)(1/cosx) = tanxsecx
--------------
y = cosecx = 1/sinx

Using Implicit Differentiation:

y = 1/sinx
ysinx = 1
dy/dx.sinx + ycosx = 0
dy/dx.sinx = - ycosx
dy/dx = - ycosx/sinx Now sub in y = 1/sinx ...
dy/dx = -(1/sinx)(cosx/sinx)

dy/dx = -cosec[x]cot[x]
--------------
cosecx + secx = 1/sinx + 1/cosx

1/sinx + 1/cosx = (cosx + sinx) / sin[x]cos[x]

Now use identity 2sin[x]cos[x] = sin[2x]

sin[x]cos[x] = ½sin[2x] , sub this result into equation:

.... (cosx + sinx) / ½sin[2x]
.... 2(cosx + sinx) / sin[2x]

whats implicit differentiation ?
Reply 8
im slightly confused,
im not sure which questions have been done?
Reply 9
lil_kk
whats implicit differentiation ?


It is something you should learn in c4. It basically means when you differentiate a function that has more then one variable, your used example y=x then dy/dx=1 but what if you had x²+y²+x+y=1 you cant just say dy/dx=2x+2y+2 since you should only be differentiating with respect to x not y.

This is hard to explain if you have Heinmann c4 book go to page 33 and read examples.
Reply 10
mala2k
It is something you should learn in c4. It basically means when your differentiation a function that has more then one variable your used to differentation y=x dy/dx=1 but what if you had x²+y²+x+y=1 you cant just say dy/dx=2x+2y+2 since you should only be differentiating with respect to x not y. This is hard to explain if you have Heinmann c4 book go to page 33 and read examples.

o..i see
i havent got that far yet lol, im starting partial fractions :smile:
--------------
mala2k
It is something you should learn in c4. It basically means when you differentiate a function that has more then one variable, your used example y=x then dy/dx=1 but what if you had x²+y²+x+y=1 you cant just say dy/dx=2x+2y+2 since you should only be differentiating with respect to x not y.

This is hard to explain if you have Heinmann c4 book go to page 33 and read examples.


btw is there another way of answering these questions
Reply 11
btw is there another way of answering these questions


Sorry apart from that first part , I would have used it as well. Is this supposed to be c3 differentiation?
Reply 12
mala2k
Sorry apart from that first part , I would have used it as well. Is this supposed to be c3 differentiation?

yup
Reply 13
fallen_star
yup


:redface: Sorry what am I on about , I just looked at the question he used implicit differentiation on you can do it another way:

y=sec x dy/dx=secxtanx
sec x = 1/cos x
.:.y=1/cos x

y=(cosx)-1
.dy/dx= -(cosx)-2 (-sinx)
........=sinx/cos2 x
........=(1/cosx)(sinx/cosx)
........=sec x tan x

Same for the other part:
y=cosec x dy/dx=-cosecxcotx
cosec x = 1/sin x
.:. y=1/sin x

y=(sinx)-1
.dy/dx= -(sinx)-2 (cosx)
........=-cosx/sin2 x
........=(-1/sinx)(cosx/sinx)
........=-cosec xcot x

Thats just normal trig differentiation you should learn in c3.
Reply 14
Same for the other part:
y=cosec x dy/dx=-cosecxcotx
cosec x = 1/sin x
y=1/sin x

y=(sinx)-1
.dy/dx= -(sinx)-2 (cosx)
........=-cosx/sin2 x
........=(-1/sinx)(cosx/sinx)
........=-cosec xcot x
Reply 15
mala2k
:redface: Sorry what am I on about , I just looked at the question he used implicit differentiation on you can do it another way:

y=sec x dy/dx=secxtanx
sec x = 1/cos x
.:.y=1/cos x

y=(cosx)-1
.dy/dx= -(cosx)-2 (-sinx)
........=sinx/cos2 x
........=(1/cosx)(sinx/cosx)
........=sec x tan x

Same for the other part:
y=cosec x dy/dx=-cosecxcotx
cosec x = 1/sin x
.:. y=1/sin x

y=(sinx)-1
.dy/dx= -(sinx)-2 (cosx)
........=-cosx/sin2 x
........=(-1/sinx)(cosx/sinx)
........=-cosec xcot x

Thats just normal trig differentiation you should learn in c3.

o.. okie thanks soo much for your help