Solve 5x^4y = 16 - ScanMath

Solve the equation

Solve for x

Solve for y

x = - \frac{2 4 125 y ^{3}}{5 ∣ y ∣}, y \neq = 0 x = \frac{2 4 125 y ^{3}}{5 ∣ y ∣}, y \neq = 0

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Testing for symmetry about the origin

Testing for symmetry about the x-axis

Testing for symmetry about the y-axis

Not symmetry with respect to the origin

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r = \frac{5 16}{5 5 cos ^{4} ( θ ) sin ( θ )}

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Find the derivative with respect to x

Find the derivative with respect to y

\frac{d y}{d x} = - \frac{4 y}{x}

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Find the second derivative with respect to x

Find the second derivative with respect to y

\frac{d ^{2} y}{d x ^{2}} = \frac{20 y}{x ^{2}}

Calculate

5 x^{4} y = 16

Take the derivative of both sides

\frac{d}{d x} (5 x^{4} y) = \frac{d}{d x} (16)

Calculate the derivative

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20 x^{3} y + 5 x^{4} \frac{d y}{d x} = \frac{d}{d x} (16)

Calculate the derivative

20 x^{3} y + 5 x^{4} \frac{d y}{d x} = 0

Move the expression to the right-hand side and change its sign

5 x^{4} \frac{d y}{d x} = 0 - 20 x^{3} y

Removing 0 doesn't change the value,so remove it from the expression

5 x^{4} \frac{d y}{d x} = - 20 x^{3} y

Divide both sides

\frac{5 x ^{4} \frac{d y}{d x}}{5 x ^{4}} = \frac{- 20 x ^{3} y}{5 x ^{4}}

Divide the numbers

\frac{d y}{d x} = \frac{- 20 x ^{3} y}{5 x ^{4}}

Divide the numbers

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\frac{d y}{d x} = - \frac{4 y}{x}

Take the derivative of both sides

\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (- \frac{4 y}{x})

Calculate the derivative

\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} (- \frac{4 y}{x})

Use differentiation rules

\frac{d ^{2} y}{d x ^{2}} = - \frac{\frac{d}{d x} ( 4 y ) \times x - 4 y \times \frac{d}{d x} ( x )}{x ^{2}}

Calculate the derivative

More Steps

\frac{d ^{2} y}{d x ^{2}} = - \frac{4 \frac{d y}{d x} \times x - 4 y \times \frac{d}{d x} ( x )}{x ^{2}}

Use \frac{d}{d x} x^{n} = n x^{n - 1} to find derivative

\frac{d ^{2} y}{d x ^{2}} = - \frac{4 \frac{d y}{d x} \times x - 4 y \times 1}{x ^{2}}

Use the commutative property to reorder the terms

\frac{d ^{2} y}{d x ^{2}} = - \frac{4 x \frac{d y}{d x} - 4 y \times 1}{x ^{2}}

Any expression multiplied by 1 remains the same

\frac{d ^{2} y}{d x ^{2}} = - \frac{4 x \frac{d y}{d x} - 4 y}{x ^{2}}

Use equation \frac{d y}{d x} = - \frac{4 y}{x} to substitute

\frac{d ^{2} y}{d x ^{2}} = - \frac{4 x ( - \frac{4 y}{x} ) - 4 y}{x ^{2}}

Solution

More Steps

\frac{d ^{2} y}{d x ^{2}} = \frac{20 y}{x ^{2}}

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