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

The answer to (\partial)/(\partial x)(ln(x^4+y^2)) is (4x^3)/(x^4+y^2)

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

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

\frac{4 x ^{3}}{x ^{4} + y ^{2}}

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (x^{4} + y^{2}) = 4 x^{3}

= \frac{1}{x ^{4} + y ^{2}} \cdot 4 x^{3}

Simplify

\frac{1}{x ^{4} + y ^{2}} \cdot 4 x^{3} : \frac{4 x ^{3}}{x ^{4} + y ^{2}}

= \frac{4 x ^{3}}{x ^{4} + y ^{2}}

y^{^{'}} = 7 x + y, y (0) = 8

y = x^{3}, y = 3 x + 2

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

\int_{0}^{1} \frac{1}{t + 1} d t

\frac{d y}{d x}, x y = 36

What is (\partial)/(\partial x)(ln(x^4+y^2)) ?
The answer to (\partial)/(\partial x)(ln(x^4+y^2)) is (4x^3)/(x^4+y^2)