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

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

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

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

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derivative of

y = (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}

\frac{d}{d x} (\frac{5}{x ^{7}})

- 3 e^{4 t}

\int_{1}^{3} \frac{e ^{x}}{e ^{x} - 1} d x

\int 3 - t e^{- t} d t

\int \frac{1 - 3 x}{2 1 - x} d x

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