- #1

francisg3

- 32

- 0

This is what I have done so far:

**f(x)=(x cosx- sinx)/(x-sinx)****1st trig function**f(x)=x cosx

f'(x) =cosx-x sinx

f''(x) =-sin〖x-sinx-x cosx 〗

f'''(x) =-cos〖x-cosx-cosx 〗+x sinx f'''(0) =-3

**3rd Maclaurin Polynomial**-3/3! x^3

**2nd trig function**f(x)=-sinx

f'(x)=-cosx

f''(x) =sinx

f'''(x) =cosx = f'''(0) =1

**3rd Maclaurin Polynomial**(1/3!) x3

**Replacing those values in the original function I get:**

(-3/3! x^3+1/3! x^3)/(1+1/3! x^3 )

I should get an answer around -2 I believe however this is not the case. Any help would be appreciated.

Thanks!