What is the derivative of e^{ln(x^2+1}) ?

The derivative of e^{ln(x^2+1}) is 2x

What is the first derivative of e^{ln(x^2+1}) ?

The first derivative of e^{ln(x^2+1}) is 2x

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

2 x

Solution steps

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

Simplify

e^{l n (x^{2} + 1)} : x^{2} + 1

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

Apply the Sum/Difference Rule:

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

= \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

= 2 x + 0

Simplify

= 2 x

y = x^{3} - 12 x, (1, - 11)

- x + \frac{9}{x} + 1

\frac{\partial}{\partial x} (x - 8 y + 3 x y)

y = 9 x^{2} - x^{3}, (2, 28)

\int \frac{( cos ( 3 x ) )}{1 + sin ( 3 x )} d x

What is the derivative of e^{ln(x^2+1}) ?
The derivative of e^{ln(x^2+1}) is 2x
What is the first derivative of e^{ln(x^2+1}) ?
The first derivative of e^{ln(x^2+1}) is 2x