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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 4 lim (x^{2} - 6 x + 8)

n = 1 \sum \infty (- 1)^{n} \cdot n

x \to 2 + lim (\frac{3}{2 - x})

\int \frac{1}{3 2 - x} d x

6 (1 - e^{- 5 t})

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