What is (\partial)/(\partial x)(ln(8x^2+y^2+3)) ?

The answer to (\partial)/(\partial x)(ln(8x^2+y^2+3)) is (16x)/(8x^2+y^2+3)

Popular >

(\partial)/(\partial x)(ln(8x^2+y^2+3))

\frac{\partial}{\partial x} (ln (8 x^{2} + y^{2} + 3))

\frac{16 x}{8 x ^{2} + y ^{2} + 3}

Solution steps

\frac{\partial}{\partial x} (ln (8 x^{2} + y^{2} + 3))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{8 x ^{2} + y ^{2} + 3} \frac{\partial}{\partial x} (8 x^{2} + y^{2} + 3)

= \frac{1}{8 x ^{2} + y ^{2} + 3} \frac{\partial}{\partial x} (8 x^{2} + y^{2} + 3)

\frac{\partial}{\partial x} (8 x^{2} + y^{2} + 3) = 16 x

= \frac{1}{8 x ^{2} + y ^{2} + 3} \cdot 16 x

Simplify

\frac{1}{8 x ^{2} + y ^{2} + 3} \cdot 16 x : \frac{16 x}{8 x ^{2} + y ^{2} + 3}

= \frac{16 x}{8 x ^{2} + y ^{2} + 3}

\frac{\partial}{\partial y} (15 x^{2} - 9 x y + 25 y^{2})

p (t) = 12 t e^{- \frac{t}{6}}

\int \frac{x + 1}{x ^{4} - 4 x ^{2}} d x

\int (\frac{y ^{2}}{2} - y tan (x)) d x

\frac{9}{s ( s ^{2} + 9 s + 9 )}

What is (\partial)/(\partial x)(ln(8x^2+y^2+3)) ?
The answer to (\partial)/(\partial x)(ln(8x^2+y^2+3)) is (16x)/(8x^2+y^2+3)