Hi Guys given the following question
$$y'(x) = \ln(x+y), x_0 = 1, y_0=1$$
Therefore using the taylor series method
$$y = y_0 + (x-x_0)y'_0+\frac{(x^2-2x+1)(y''_0)}{2}+\frac{x^3-3x^2+3x-1}{6}(y'''_0)$$
Therefore
$$y'(x) = \ln(x+y) = \ln x+\ln y$$
$$y''(x) = \frac{d}{dx}ln(x+y)$$
$$\frac{df}{dx} = y = ln(u), u = x+y$$
$$\frac{dy}{dx} = \frac{1}{x+y}$$
$$\frac{df}{dy} = \frac{y'}{x+y}$$
hence the we get the following
$$y''(x) = \frac{1}{x+y}+\frac{1}{x+y}y'$$
$$y''(x) = \frac{1+y'}{x+y}$$
$$y'''(x)= \frac{1+y'}{x+y}$$
$$let u = 1+y', v = x+y$$
$$\frac{du}{dx} = y'', \frac{dv}{dx} = 1+y'$$
Placing into the formula
$$\frac{y''(x+y)-(1+y')(1+y')}{(x+y)^2}$$
$$y'(x) = \ln(x+y), x_0 = 1, y_0=1$$ The second derivative of y is: $$y''=\dfrac {1+y'}{x+y}$$ $$y''=\dfrac {1+\ln |x+y|}{x+y}$$ You can't write: $$\ln (x+y)=\ln (x)+\ln(y)$$ Don't confuse with: $$\ln (xy)=\ln (x)+\ln(y)$$
You need to differentiate a log function so: $$(\ln (x+y))'=\frac 1{x+y}(x+y)'$$ And $(x+y)'=1+y'$. Or for any function $f(x)$: $$(\ln (f(x)))'=\frac 1{f(x)}(f(x))'$$
If you prefer substitution and chain rule : $u=x+y$ you have : $$y'= \ln u $$ $$y''=\dfrac {d \ln u}{du}\dfrac {du}{dx}=\frac 1 u u'$$ and $u'=1+y'$ $$y''=\dfrac {1+y'}{x+y}$$ $$y''=\dfrac {1+\ln (x+y)}{x+y}$$ Use quotient rule for the third derivative: $$y^{(3)}=-\dfrac {y'(1+y')}{(x+y)^2}$$ $$y^{(3)}=-\dfrac {(1+\ln (x+y))\ln (x+y)}{(x+y)^2}$$