What is expand (d^2)/(d^2x)(y^4+3y-4x^3=5x+1) ?

The solution to expand (d^2)/(d^2x)(y^4+3y-4x^3=5x+1) is (y^4)/x+(3y)/x-4x^2=5x+1/x

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expand (d^2)/(d^2x)(y^4+3y-4x^3=5x+1)

expand

\frac{d ^{2}}{d ^{2} x} (y^{4} + 3 y - 4 x^{3} = 5 x + 1)

\frac{y ^{4}}{x} + \frac{3 y}{x} - 4 x^{2} = 5 x + \frac{1}{x}

Solution steps

\frac{d ^{2}}{d ^{2} x} (y^{4} + 3 y - 4 x^{3} = 5 x + 1)

= \frac{d ^{2}}{d ^{2} x} (y^{4} + 3 y - 4 \cdot 5 = x^{3} x + 1)

Distribute parentheses

= \frac{d ^{2}}{d ^{2} x} y^{4} + \frac{d ^{2}}{d ^{2} x} \cdot 3 y + \frac{d ^{2}}{d ^{2} x} (- 4 x^{3} = 5 x) + \frac{d ^{2}}{d ^{2} x} \cdot 1

Apply minus-plus rules

+ (- a) = - a

\frac{d ^{2}}{d ^{2} x} y^{4} + 3 \cdot \frac{d ^{2}}{d ^{2} x} y - 4 \cdot \frac{d ^{2}}{d ^{2} x} x^{3} = 5 x + 1 \cdot \frac{d ^{2}}{d ^{2} x}

Simplify

\frac{d ^{2}}{d ^{2} x} y^{4} + 3 \cdot \frac{d ^{2}}{d ^{2} x} y - 4 \cdot \frac{d ^{2}}{d ^{2} x} x^{3} = 5 x + 1 \cdot \frac{d ^{2}}{d ^{2} x} : \frac{y ^{4}}{x} + \frac{3 y}{x} - 4 x^{2} = 5 x + \frac{1}{x}

\frac{y ^{4}}{x} + \frac{3 y}{x} - 4 x^{2} = 5 x + \frac{1}{x}

(y^{2} - x^{2}) \frac{d y}{d x}

\frac{( - 9 x + 8 ) ( x - 8 )}{( - 9 x - 8 ) x}

6 (\frac{4 x ^{3}}{3} - 2 x^{2} + x)

\frac{( 4 x - 1 ) ^{2}}{16}

C (- 3, - 5)

What is expand (d^2)/(d^2x)(y^4+3y-4x^3=5x+1) ?
The solution to expand (d^2)/(d^2x)(y^4+3y-4x^3=5x+1) is (y^4)/x+(3y)/x-4x^2=5x+1/x