What is (\partial)/(\partial x)(ln(9x^2+4y^2+5)) ?

The answer to (\partial)/(\partial x)(ln(9x^2+4y^2+5)) is (18x)/(9x^2+4y^2+5)

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(\partial)/(\partial x)(ln(9x^2+4y^2+5))

\frac{\partial}{\partial x} (ln (9 x^{2} + 4 y^{2} + 5))

\frac{18 x}{9 x ^{2} + 4 y ^{2} + 5}

Solution steps

\frac{\partial}{\partial x} (ln (9 x^{2} + 4 y^{2} + 5))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{9 x ^{2} + 4 y ^{2} + 5} \frac{\partial}{\partial x} (9 x^{2} + 4 y^{2} + 5)

= \frac{1}{9 x ^{2} + 4 y ^{2} + 5} \frac{\partial}{\partial x} (9 x^{2} + 4 y^{2} + 5)

\frac{\partial}{\partial x} (9 x^{2} + 4 y^{2} + 5) = 18 x

= \frac{1}{9 x ^{2} + 4 y ^{2} + 5} \cdot 18 x

Simplify

\frac{1}{9 x ^{2} + 4 y ^{2} + 5} \cdot 18 x : \frac{18 x}{9 x ^{2} + 4 y ^{2} + 5}

= \frac{18 x}{9 x ^{2} + 4 y ^{2} + 5}

\frac{\partial}{\partial y} (x^{3} - 2 x y + 15)

\int x arcsec (2 x) d x

\int \frac{1}{e ^{- v} + 1} d v

\int x^{3} \cdot sin (2 x) d x

y = \frac{5}{x ^{4}} - \frac{7}{x}

What is (\partial)/(\partial x)(ln(9x^2+4y^2+5)) ?
The answer to (\partial)/(\partial x)(ln(9x^2+4y^2+5)) is (18x)/(9x^2+4y^2+5)