What is the derivative of y=ln(x^3+4) ?

The derivative of y=ln(x^3+4) is (3x^2)/(x^3+4)

What is the first derivative of y=ln(x^3+4) ?

The first derivative of y=ln(x^3+4) is (3x^2)/(x^3+4)

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derivative of

y = ln (x^{3} + 4)

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

Solution steps

\frac{d}{d x} (ln (x^{3} + 4))

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (6 x - 2 y)

y^{^{'}} = 4 x + y, y (0) = 7

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 1

\frac{2}{5 x + 3}

\int \frac{csc ^{2} ( x )}{cot ^{3} ( x )} d x

What is the derivative of y=ln(x^3+4) ?
The derivative of y=ln(x^3+4) is (3x^2)/(x^3+4)
What is the first derivative of y=ln(x^3+4) ?
The first derivative of y=ln(x^3+4) is (3x^2)/(x^3+4)