What is the derivative of In(x^2+1) ?

The derivative of In(x^2+1) is 2nIx

What is the first derivative of In(x^2+1) ?

The first derivative of In(x^2+1) is 2nIx

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

2 n I x

Solution steps

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

Treat

n, I

as a constant

Take the constant out:

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

= I n \frac{d}{d x} (x^{2} + 1)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= I n (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (1))

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

\frac{d}{d x} (1) = 0

= I n (2 x + 0)

Simplify

= 2 n I x

\frac{\partial}{\partial z} (2 x z^{2})

x \to \infty lim (x^{a})

y = 2 x - 2 x^{2}, at x = - 1.5

y = (x + 1)^{x^{2}}, (0, 1)

\int \frac{3}{3 x + 2} d x

What is the derivative of In(x^2+1) ?
The derivative of In(x^2+1) is 2nIx
What is the first derivative of In(x^2+1) ?
The first derivative of In(x^2+1) is 2nIx