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

The derivative of in(x^2+1) is 2inx

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

The first derivative of in(x^2+1) is 2inx

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

2 in x

Solution steps

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

Simplify

in (x^{2} + 1) : i (n + n x^{2})

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

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

\frac{d n}{d x} = 0

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

= i (0 + 2 n x)

Simplify

= 2 in x

\frac{d y}{d x} = y^{2} + 1

\frac{d}{d x} (e^{\frac{- x ^{2}}{8}})

\int (ln (x))^{4} \frac{1}{x} d x

\int_{3}^{4} \frac{x}{x ^{2} - 1} d x

\int \frac{1}{( x - 3 )} d x

What is the derivative of in(x^2+1) ?
The derivative of in(x^2+1) is 2inx
What is the first derivative of in(x^2+1) ?
The first derivative of in(x^2+1) is 2inx