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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

y = 2 x^{2} + 5 x + 6, (- 2, 0)

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

\int 4 \frac{cos ( x )}{sin ^{2} ( x )} d x

2 x^{2} - 7 x + 5

\frac{\partial}{\partial x} (C e^{2 x})

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