Question

to find nth derivative of x/(x+1)^4

Answer 1

1-3x/(1+x)
$1 - 3x \div {(x + 1)}^{5}$

Answer 2

Answer:

To find the nth derivative of the function $y=\frac{x}{(x+1)^{4}}=\frac{x+1-1}{(x+1)^{4}}=\frac{1}{(x+1)^{3}}-\frac{1}{(x+1)^{4}}$

Step-by-step explanation:

So, the function

$y=(x+1)^{-3}-(x+1)^{-4}$

Therefore, the derivatives are:

$y_{1} &=-3(x+1)^{-3-1}+4(x+1)^{-4-1} \\\\=-3(x+1)^{-4}+4(x+1)^{-5} \\\\y_{2} &=(-3)(-4)(x+1)^{-5}-(-4)(-5)(x+1)^{-6} \\\\=\frac{4 !(-1)^{2}}{2}(x+1)^{-5}-\frac{5 !(-1)^{2}}{3 !}(x+1)^{-6} \\\\y_{3} &=\frac{5 !(-1)^{3}(x+1)^{-6}}{2}-\frac{6 !(-1)^{3}(x+1)^{-7}}{6}\\\\y_{\mathrm{n}}=\frac{(\mathrm{n}+2) !(-1)^{\mathrm{n}}(\mathrm{x}+1)^{-\mathrm{n}-3}}{2}-\frac{(\mathrm{n}+3) !(\mathrm{x}+1)^{-\mathrm{n}-4}}${6}

So, the function

$y=(x+1)^{-3}-(x+1)^{-4}$

Therefore, the derivatives are:

$y_{1} &=-3(x+1)^{-3-1}+4(x+1)^{-4-1} \\\\&=-3(x+1)^{-4}+4(x+1)^{-5} \\\\y_{2} &=(-3)(-4)(x+1)^{-5}-(-4)(-5)(x+1)^{-6} \\\\&=\frac{4 !(-1)^{2}}{2}(x+1)^{-5}-\frac{5 !(-1)^{2}}{3 !}(x+1)^{-6} \\\\y_{3} &=\frac{5 !(-1)^{3}(x+1)^{-6}}{2}-\frac{6 !(-1)^{3}(x+1)^{-7}}{6}\\y_{\mathrm{n}}=\frac{(\mathrm{n}+2) !(-1)^{\mathrm{n}}(\mathrm{x}+1)^{-\mathrm{n}-3}}{2}-\frac{(\mathrm{n}+3) !(\mathrm{x}+1)^{-\mathrm{n}-4}}{6}$