What is (\partial)/(\partial x)(ln(5x^2+y^2+9)) ?

The answer to (\partial)/(\partial x)(ln(5x^2+y^2+9)) is (10x)/(5x^2+y^2+9)

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(\partial)/(\partial x)(ln(5x^2+y^2+9))

\frac{\partial}{\partial x} (ln (5 x^{2} + y^{2} + 9))

\frac{10 x}{5 x ^{2} + y ^{2} + 9}

Solution steps

\frac{\partial}{\partial x} (ln (5 x^{2} + y^{2} + 9))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{5 x ^{2} + y ^{2} + 9} \frac{\partial}{\partial x} (5 x^{2} + y^{2} + 9)

= \frac{1}{5 x ^{2} + y ^{2} + 9} \frac{\partial}{\partial x} (5 x^{2} + y^{2} + 9)

\frac{\partial}{\partial x} (5 x^{2} + y^{2} + 9) = 10 x

= \frac{1}{5 x ^{2} + y ^{2} + 9} \cdot 10 x

Simplify

\frac{1}{5 x ^{2} + y ^{2} + 9} \cdot 10 x : \frac{10 x}{5 x ^{2} + y ^{2} + 9}

= \frac{10 x}{5 x ^{2} + y ^{2} + 9}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = 0

f (x) = \frac{22}{( 1 - x ) ^{3}}

y = x^{2} - 2 x, (3, 3)

\frac{d y}{d x} = \frac{x ^{2}}{9} + \frac{7}{9}

\int \frac{( x ^{2} - 4 x + 4 )}{x - 2} d x

What is (\partial)/(\partial x)(ln(5x^2+y^2+9)) ?
The answer to (\partial)/(\partial x)(ln(5x^2+y^2+9)) is (10x)/(5x^2+y^2+9)