What is (\partial)/(\partial x)(ln(1+3x^2y^3)) ?

The answer to (\partial)/(\partial x)(ln(1+3x^2y^3)) is (6xy^3)/(1+3x^2y^3)

Popular >

(\partial)/(\partial x)(ln(1+3x^2y^3))

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

\frac{6 x y ^{3}}{1 + 3 x ^{2} y ^{3}}

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (1 + 3 x^{2} y^{3}) = 6 y^{3} x

= \frac{1}{1 + 3 x ^{2} y ^{3}} \cdot 6 y^{3} x

Simplify

\frac{1}{1 + 3 x ^{2} y ^{3}} \cdot 6 y^{3} x : \frac{6 x y ^{3}}{1 + 3 x ^{2} y ^{3}}

= \frac{6 x y ^{3}}{1 + 3 x ^{2} y ^{3}}

x \to 3 lim (\frac{x ^{2} - 9}{x + 3})

\int 8 x cos^{2} (x) d x

y^{^{'}} + sin (t) y = - sin (t)

\frac{\partial}{\partial x} (arcsin (10 x yz))

y = \frac{1 + x}{1 + e ^{x}}, (0, \frac{1}{2})

What is (\partial)/(\partial x)(ln(1+3x^2y^3)) ?
The answer to (\partial)/(\partial x)(ln(1+3x^2y^3)) is (6xy^3)/(1+3x^2y^3)