What is the derivative of ln(x^2+y^4) ?

The derivative of ln(x^2+y^4) is (2x+4y^3\frac{dy)/(dx)}{x^2+y^4}

What is the first derivative of ln(x^2+y^4) ?

The first derivative of ln(x^2+y^4) is (2x+4y^3\frac{dy)/(dx)}{x^2+y^4}

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

4 cos (5 t)

y^{^{'}} + y = 3

\frac{5 x}{x - 2}

y^{^{'}} = 6 x 9 - y^{2}

y^{^{'}} = 1 + x + y^{2} + x y^{2}

What is the derivative of ln(x^2+y^4) ?
The derivative of ln(x^2+y^4) is (2x+4y^3\frac{dy)/(dx)}{x^2+y^4}
What is the first derivative of ln(x^2+y^4) ?
The first derivative of ln(x^2+y^4) is (2x+4y^3\frac{dy)/(dx)}{x^2+y^4}