What is (\partial)/(\partial x)((ln(2x^2+5))/(y^2)) ?

The answer to (\partial)/(\partial x)((ln(2x^2+5))/(y^2)) is (4x)/(y^2(2x^2+5))

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(\partial)/(\partial x)((ln(2x^2+5))/(y^2))

\frac{\partial}{\partial x} (\frac{ln ( 2 x ^{2} + 5 )}{y ^{2}})

\frac{4 x}{y ^{2} ( 2 x ^{2} + 5 )}

Solution steps

\frac{\partial}{\partial x} (\frac{ln ( 2 x ^{2} + 5 )}{y ^{2}})

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{y ^{2}} \frac{\partial}{\partial x} (ln (2 x^{2} + 5))

Apply the chain rule:

\frac{1}{2 x ^{2} + 5} \frac{\partial}{\partial x} (2 x^{2} + 5)

= \frac{1}{2 x ^{2} + 5} \frac{\partial}{\partial x} (2 x^{2} + 5)

\frac{\partial}{\partial x} (2 x^{2} + 5) = 4 x

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

Simplify

\frac{1}{y ^{2}} \cdot \frac{1}{2 x ^{2} + 5} \cdot 4 x : \frac{4 x}{y ^{2} ( 2 x ^{2} + 5 )}

= \frac{4 x}{y ^{2} ( 2 x ^{2} + 5 )}

\frac{\partial}{\partial y} (\frac{y ^{2}}{2})

\frac{d}{d x} (\frac{x ^{5}}{1 - x ^{4}})

\frac{d}{d x} ((x^{2} + 2)^{3})

x \to \infty lim (5 - (- 3))

\int \frac{1}{( 3 - 4 x )} d x

What is (\partial)/(\partial x)((ln(2x^2+5))/(y^2)) ?
The answer to (\partial)/(\partial x)((ln(2x^2+5))/(y^2)) is (4x)/(y^2(2x^2+5))