well i have p3 tomorrow so some help is required

[Tobe]

y=log(base10)x
x=10^y

show that x may be expressed in the form e^yln10

i have solutions i just need to understand why/how you substitute e in

and also

differentiate xe^(3x)

can someone explain the solution to me

and while your at that use the derivative of cos to prove d/dx(secx)=secxtanx

way im told to solve this is using d/dx(1/cosx)= sinx/cosx^2

but like how does d/dx of 1/cosx= sinx/cosx2 and how does that =secxtanx

cheers

[MrL_JaKiri]

y = lg x

y = ln x/ln 10

yln10 = ln x

e ^ y ln 10 = x

as for the 2nd

use the product rule.

d (xe^3x)/dx = dx/dx * e^3x + x * d e^3x/dx

= e^3x(1 + 3x)

as for the third

d (1/cosx)/dx = - (d cos x/dx) / cos^2x

(by d(f(x)^n)/dx = f'(x) * n(f(x))^n-1)

= - (-sin x)/cos^2x

= sin/cos * 1/cos

= tan x sec x

[Freakozoid]

i have Stats on wednesday.

OH NO WHAT DO I DO?

[MrL_JaKiri]

Originally posted by Freakozoid
i have Stats on wednesday.

OH NO WHAT DO I DO?

Pass.

It's easy.

[CoD]

I have P3 on friday 13th
I recon i need about half marks to finish up with an A
It looks hard tho, way harder than p1,p2,m1,m2,s1

[Deathjam]

i don't have a ****ing clue what ur on about..


but lookie i can type crap too

f = 5 + w

f * w = g / 5 *2

g > 6 - wq * p / 0

8 = (6^r)+(b+s)

[MrL_JaKiri]

Originally posted by Deathjam
i don't have a ****ing clue what ur on about..


but lookie i can type crap too

f = 5 + w

f * w = g / 5 *2

g > 6 - wq * p / 0

8 = (6^r)+(b+s)

There's nothing like a good maths post, and this is nothing like a good maths post*!


*This joke sponsored by the 'let DM relive his youth' campaign.