<s>[INST] Simplify $ \frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}} $ Please help me find the sum
$$
\frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}}
$$
 [/INST] Note that each term is of the form
$$\dfrac{df_k/dx}{f_k} = \dfrac{d(\log(f_k)))}{dx}$$
Hence,
$$\sum_{k=0}^n \dfrac{df_k/dx}{f} = \sum_{k=1}^n \dfrac{d(\log(f_k)))}{dx} = \dfrac{d(\sum_{k=0}^n \log(f_k)))}{dx} = \dfrac{d(\log(f_0 \cdot f_1 \cdot f_2 \cdot f_3 \cdots f_n))}{dx}$$In your case, $f_0 f_1 f_2 \cdots f_n$ reduces to give a nice short expression, which can be easily differentiated. 
Move the cursor over the gray area for a complete answer.

In your case, $f_0 = (x-y)$, $f_n = x^{2^{n-1}} + y^{2^{n-1}}$ for $n > 0$. Hence, $$f_0 f_1 f_2 \cdots f_n = (x-y)(x+y)(x^2+y^2)\cdots (x^{2^{n-1}} + y^{2^{n-1}}) = x^{2^n} - y^{2^n}$$ Hence, $$\dfrac{d(\log(f_0 \cdot f_1 \cdot f_2 \cdot f_3 \cdots f_n))}{dx} = \dfrac{2^n x^{2^n-1}}{x^{2^n}-y^{2^n}}$$ In your case, $n=5$.

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