Question

can you help me with this for my exercises:

I need to to show that:
g(x)= (e(-3x+2sinx)/6)
----------------
(3+2cosx)

has derivative

g'(x)= - (4(sin^2)x -12sin x +5) (e(-3x+2sinx)/6)
-----------------------------------
6(3+2cosx)^2

Answer 1

this is g ?

\[g(x) = \frac{e^{\frac{1}{6}(-3x +2sinx)}}{3+2cosx}\]

Answer 2

let let f(x) be a function of x

if we have a function of a function

eg g(f(x))

when we differentiate we get
g'(f(x)).f'(x)

Answer 3

if we have two functions u(x) and v(x)

multiplied together:

u(x)v(x)

when we differentiate we get

u'(x)v(x) + v'(x)u(x)

Answer 4

lets take a look at your function g(x)

Answer 5

we could split it up:

\[g(x) = \frac{1}{3+2cosx} \times e^{\frac{1}{6}(3x + 2sinx)} = u(x) \times v(x)\]

\[u(x) = \frac{1}{3+2cosx} \]

\[v(x) =e^{\frac{1}{6}(3x + 2sinx)}\]

so to differentiate we need:

\[u'(x)v(x) +v'(x)u(x)\]

\[( \frac{1}{3+2cosx} \times \frac{d}{dx} (e^{\frac{1}{6}(3x + 2sinx)}) )+ (e^{\frac{1}{6}(3x + 2sinx)} \times \frac{d}{dx} (\frac{1}{3+2cosx}))\]

Answer 6

ok so quotient rule says


if h(x) = f(x)/g(x)

then h'(x) = (f'(x)g(x) - g'(x)f(x))/(g(x))^2

Answer 7

that is how we differentiate u(x)

to differentiate v(x) we need to use the chain rule, which i started with :)

Answer 8

Could you help do it with the quotient rule?

Answer 9

Ok thanks.

Answer 10

Appreciate your help