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

The derivative of x^{ln(x^2+1}) is ((2xln(x)}{x^2+1}+\frac{ln(x^2+1))/x)(x^2+1)^{ln(x)}

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

The first derivative of x^{ln(x^2+1}) is ((2xln(x)}{x^2+1}+\frac{ln(x^2+1))/x)(x^2+1)^{ln(x)}

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

\frac{d ^{2}}{d x ^{2}} (e^{s i n (x)})

\frac{d}{d x} ((sin (y)) x^{2})

x \to 5 lim (h (x))

y^{^{'}} + 5 (tan (5 x)) y = - 8 (cos (5 x))

1 - cos (7 t)

What is the derivative of x^{ln(x^2+1}) ?
The derivative of x^{ln(x^2+1}) is ((2xln(x)}{x^2+1}+\frac{ln(x^2+1))/x)(x^2+1)^{ln(x)}
What is the first derivative of x^{ln(x^2+1}) ?
The first derivative of x^{ln(x^2+1}) is ((2xln(x)}{x^2+1}+\frac{ln(x^2+1))/x)(x^2+1)^{ln(x)}