What is the derivative of (1+x)^n ?

The derivative of (1+x)^n is n(1+x)^{n-1}

What is the first derivative of (1+x)^n ?

The first derivative of (1+x)^n is n(1+x)^{n-1}

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derivative of

(1 + x)^{n}

n (1 + x)^{n - 1}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

= n (1 + x)^{n - 1}

\frac{d}{d x} (x^{2^{x}})

y^{^{''}} - (\frac{2}{t}) y^{^{'}} + (\frac{2}{t ^{2}}) y = \frac{2 t}{1 + t ^{2}}

\int x^{9} x^{5} + 4 d x

\frac{d}{d y} (y^{- 2})

x - 1

What is the derivative of (1+x)^n ?
The derivative of (1+x)^n is n(1+x)^{n-1}
What is the first derivative of (1+x)^n ?
The first derivative of (1+x)^n is n(1+x)^{n-1}