Examine's 2u horrors (1 Viewer)

[ravana94]

Hey guys I was wondering if you couldo this trig differentiation( because i couldn't )

11a) show that 1/2(sin(m+n)x+sin(m-n)x) = sinmx cosnx. (this is the easy bit)
b) Hence, without using the product rule, find the derivative of sin mx cos nx ( )

 

[bleakarcher]

Just utilise part a). So:
d/dx[sin(mx)cos(nx)]=d/dx[(1/2)[sin(m+n)x+sin(m-n)x]=(1/2)[(m+n)cos(m+n)x+(m-n)cos(m-n)x]

 

[Carrotsticks]

Hey guys I was wondering if you couldo this trig differentiation( because i couldn't )

11a) show that 1/2(sin(m+n)x+sin(m-n)x) = sinmx cosnx. (this is the easy bit)
b) Hence, without using the product rule, find the derivative of sin mx cos nx ( )

 

[bleakarcher]

Hey Carrot, done with the videos yet? ^

 

[ravana94]

Hey thanks for your working out it really helped, time to do some more math

 

[Lemiixem]

How to differentiate:
Y = Log 1 / (2x-3)

 

[someth1ng]

How to differentiate:
Y = Log 1 / (2x-3)

Haven't done this in ages:
Let f(x)=1/(2x-1)-->f'(x)=[1/2]ln(x-1/2)
y'=f'(x)/f(x)=[(1/2)ln(x-1/2)][1/(2x-1)]=ln(x-(1/2))/4x-2)

Correct me if I'm wrong.

 

[bleakarcher]

-2/(2x-3)

 

[Carrotsticks]

How to differentiate:
Y = Log 1 / (2x-3)

log laws.

 

[someth1ng]

lol, forgot you could do that - I forgot my 2U stuff.

 

[SpiralFlex]

The question came up in Terry Lee integrals that trigonometric expansion.

Quite fun.

 

[Examine]

This week seems to be easy.

 

[Timske]

trolollolol lalalala lalalal trolololol lala hooo haha hooo haha

 

[Examine]

Need help with this question.

The roots of the quadratic equation
x^2-(2k+4)x+(k^2+3k+2)=0
are non-zero and one of the roots is twice the other root, calculate the value of k.

 

[Carrotsticks]

Need help with this question.

The roots of the quadratic equation
x^2-(2k+4)x+(k^2+3k+2)=0
are non-zero and one of the roots is twice the other root, calculate the value of k.

Let the roots be a and 2a (since one is twice the other).

Sum of roots is 2k+4, which is equal to 3a (I added the two roots).

Thus, a = 2/3 * (k+2)

P(a) = 0 since a is a root.

Solve for k.

 

[Examine]

Ahh I see, thanks.

 

[SpiralFlex]

Note: Harrians in this thread. It is important you attend all classes. Also listen and do the homework.

 

[Examine]

Note: Harrians in this thread. It is important you attend all classes. Also listen and do the homework.

Is it better to do all homework in the weekends or spread it out evenly?

 

[AAEldar]

Is it better to do all homework in the weekends or spread it out evenly?

Do it as you get it. So if you got homework today and you have maths tomorrow do it tonight.

 

[Examine]

Let the roots be a and 2a (since one is twice the other).

Sum of roots is 2k+4, which is equal to 3a (I added the two roots).

Thus, a = 2/3 * (k+2)

P(a) = 0 since a is a root.

Solve for k.

Can you clarify this step?