# Let y=ln(1 + tanx) i) Show that d^2y/dx^2 = -1 when x = 0 ii) Find the Maclaurin's series for y up to and including the term in x^2. Thanks!

*print*Print*list*Cite

### 1 Answer

i) You need to differentiate the function with respect to x such that:

`(dy)/(dx) = 1/(1+tan x)*(1+tanx)'`

`(dy)/(dx) = 1/(1+tan x)*(1/(cos^2x))`

`(dy)/(dx) = (cos x/(sin x+cos x))*(1/(cos^2x))`

`(dy)/(dx) = (1/(sin x+cos x))*(1/(cos x))`

You need to differentiate `(1/(sin x+cos x))*(1/(cos x)) ` wih respect to x such that:

`(d^2y)/(dx^2) = (-cos x+sin x)/((sin x+cos x)^2)*(1/(cos x)) + (1/(sin x+cos x))*(sin x/(cos^2x))`

You need to substitute 0 for x such that:

`(d^2y)/(dx^2)|_(x=0)= (-cos 0+sin 0)/((sin 0+cos 0)^2)*(1/(cos 0)) + (1/(sin 0+cos 0))*(sin 0/(cos^2 0))`

You need to substitute 0 for sin 0 and 1 for cos 0 such that:

`(d^2y)/(dx^2)|_(x=0)= (-1+0)/((0+1)^2)*(1/(1)) + (1/(0+1))*(0/(1))`

`(d^2y)/(dx^2)|_(x=0)= (-1)/1 + 0`

`(d^2y)/(dx^2)|_(x=0) = -1`

**Hence, evaluating `(d^2y)/(dx^2)|_(x=0)` yields `(d^2y)/(dx^2)|_(x=0) = -1.` **

ii) You need to find the Maclaurin's series such that:

`f(x) = f(0) + f'(0)*x + (f''(0)*x^2)/(2!) `

`f(x) = y = ln(1 + tan0) + (1/(sin 0+cos 0))*(1/(cos 0))*x - x^2/2`

Notice that f''(0) was evaluated to previous answer `(d^2y)/(dx^2)|_(x=0) = -1.` `y = ln1 +x - (x^2)/2`

`y =x - (x^2)/2`

**Hence, evaluating Maclaurin's series up tp `x^2` yields`y = x - (x^2)/2.` **