What is the derivative of ln(Ax^4+B) ?

The derivative of ln(Ax^4+B) is (4Ax^3)/(Ax^4+B)

What is the first derivative of ln(Ax^4+B) ?

The first derivative of ln(Ax^4+B) is (4Ax^3)/(Ax^4+B)

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derivative of

ln (A x^{4} + B)

\frac{4 A x ^{3}}{A x ^{4} + B}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{4 A x ^{3}}{A x ^{4} + B}

x \to - 3 lim (∣ x - 1 ∣)

\frac{\partial}{\partial x} (- 7 e^{- 7 y} sin (7 x))

x^{2} + y^{2} = 64, - 8, 3

y^{^{''}} + 3 y = t^{3} - 1

\frac{d}{d x} (5 e^{x} - 3 x)

What is the derivative of ln(Ax^4+B) ?
The derivative of ln(Ax^4+B) is (4Ax^3)/(Ax^4+B)
What is the first derivative of ln(Ax^4+B) ?
The first derivative of ln(Ax^4+B) is (4Ax^3)/(Ax^4+B)