What is the derivative of log_{3}(\sqrt[5]{x+2}) ?

The derivative of log_{3}(\sqrt[5]{x+2}) is 1/(5ln(3)\sqrt[5]{x+2)(x+2)^{4/5}}

What is the first derivative of log_{3}(\sqrt[5]{x+2}) ?

The first derivative of log_{3}(\sqrt[5]{x+2}) is 1/(5ln(3)\sqrt[5]{x+2)(x+2)^{4/5}}

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Solution steps

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

Apply the chain rule:

Simplify

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

\int \frac{x ^{3} + 1}{x ^{2} + 5 x + 6} d x

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

\frac{d}{d x} (750 + 12 x - 0.08 x^{7})

x^{4} + 8 x + 6

What is the derivative of log_{3}(\sqrt[5]{x+2}) ?
The derivative of log_{3}(\sqrt[5]{x+2}) is 1/(5ln(3)\sqrt[5]{x+2)(x+2)^{4/5}}
What is the first derivative of log_{3}(\sqrt[5]{x+2}) ?
The first derivative of log_{3}(\sqrt[5]{x+2}) is 1/(5ln(3)\sqrt[5]{x+2)(x+2)^{4/5}}