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

The derivative of (x^2+1^x) is (ln(x^2+1)+(2x^2)/(x^2+1))(x^2+1)^x

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

The first derivative of (x^2+1^x) is (ln(x^2+1)+(2x^2)/(x^2+1))(x^2+1)^x

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

(ln (x^{2} + 1) + \frac{2 x ^{2}}{x ^{2} + 1}) (x^{2} + 1)^{x}

Solution steps

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

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

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

Apply the chain rule:

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

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

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

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

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

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

f (x) = ln (1 + e^{x})

∣ x ∣, x^{2} - 6

y^{^{'}} + 2 y = t, y (0) = 1

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

\int arccot (10 y) d y

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