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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial t} (1 - t)

x \to 7 lim (\frac{x}{x - 3})

x \to 3 lim (\frac{sin ( 5 x )}{x})

d y = (4 x - 6) d x

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

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