What is the derivative of \sqrt[n]{ax+1} ?

The derivative of \sqrt[n]{ax+1} is (a(ax+1)^{(1-n)/n})/(n)

What is the first derivative of \sqrt[n]{ax+1} ?

The first derivative of \sqrt[n]{ax+1} is (a(ax+1)^{(1-n)/n})/(n)

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\frac{a ( a x + 1 ) ^{\frac{1 - n}{n}}}{n}

Solution steps

Apply the chain rule:

\frac{( a x + 1 ) ^{\frac{1 - n}{n}}}{n} \frac{d}{d x} (a x + 1)

= \frac{( a x + 1 ) ^{\frac{1 - n}{n}}}{n} \frac{d}{d x} (a x + 1)

\frac{d}{d x} (a x + 1) = a

= \frac{( a x + 1 ) ^{\frac{1 - n}{n}}}{n} a

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{a ( a x + 1 ) ^{\frac{1 - n}{n}}}{n}

\frac{3 e ^{- 2 s}}{4 s ^{2} + 100}

\int \frac{arctan ( 9 x )}{1 + 81 x ^{2}} d x

\frac{d y}{d x} - 2 y = x

\frac{d}{d x} y + y = 4

\int x^{2} n^{2} x d x

What is the derivative of \sqrt[n]{ax+1} ?
The derivative of \sqrt[n]{ax+1} is (a(ax+1)^{(1-n)/n})/(n)
What is the first derivative of \sqrt[n]{ax+1} ?
The first derivative of \sqrt[n]{ax+1} is (a(ax+1)^{(1-n)/n})/(n)