What is expand (y-xy'(x))^4 ?

The solution to expand (y-xy'(x))^4 is y^4-4xy^3y'(x)+6x^2y^2(y'(x))^2-4x^3y(y'(x))^3+x^4(y'(x))^4

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expand (y-xy'(x))^4

expand

(y - x y^{'} (x))^{4}

y^{4} - 4 x y^{3} y' (x) + 6 x^{2} y^{2} (y' (x))^{2} - 4 x^{3} y (y' (x))^{3} + x^{4} (y' (x))^{4}

Solution steps

(y - x y^{'} (x))^{4}

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = y, b = - x y^{'} (x)

= i = 0 \sum 4 (i 4) y^{(4 - i)} (- x y^{^{'}} (x))^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} y^{4} (- x y^{'} (x))^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} y^{3} (- x y^{'} (x))^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} y^{2} (- x y^{'} (x))^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} y^{1} (- x y^{'} (x))^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} y^{0} (- x y^{'} (x))^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} y^{4} (- x y^{'} (x))^{0} : y^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} y^{3} (- x y^{'} (x))^{1} : - 4 x y^{3} y^{'} (x)

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} y^{2} (- x y^{'} (x))^{2} : 6 x^{2} y^{2} (y^{'} (x))^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} y^{1} (- x y^{'} (x))^{3} : - 4 x^{3} y (y^{'} (x))^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} y^{0} (- x y^{'} (x))^{4} : x^{4} (y^{'} (x))^{4}

= y^{4} - 4 x y^{3} y^{^{'}} (x) + 6 x^{2} y^{2} (y^{^{'}} (x))^{2} - 4 x^{3} y (y^{^{'}} (x))^{3} + x^{4} (y^{^{'}} (x))^{4}

B (e^{x} x + 2 e^{x})

9 (z - 3 = 12 z)

\frac{1.5 ( s + 3 )}{s ( s + 1 ) ( s + 2 )}

\frac{s ( s + 11 )}{( 10 + s ) ( s + 100 )}

5 (x + 3 (x - 2) = x - 2)

What is expand (y-xy'(x))^4 ?
The solution to expand (y-xy'(x))^4 is y^4-4xy^3y'(x)+6x^2y^2(y'(x))^2-4x^3y(y'(x))^3+x^4(y'(x))^4