What is the derivative of (1+9/x)^x ?

The derivative of (1+9/x)^x is (ln(1+9/x)-9/(x+9))(1+9/x)^x

What is the first derivative of (1+9/x)^x ?

The first derivative of (1+9/x)^x is (ln(1+9/x)-9/(x+9))(1+9/x)^x

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derivative of

(1 + \frac{9}{x})^{x}

(ln (1 + \frac{9}{x}) - \frac{9}{x + 9}) (1 + \frac{9}{x})^{x}

Solution steps

\frac{d}{d x} ((1 + \frac{9}{x})^{x})

Apply exponent rule:

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

(1 + \frac{9}{x})^{x} = e^{x l n (1 + \frac{9}{x})}

= \frac{d}{d x} (e^{x l n (1 + \frac{9}{x})})

Apply the chain rule:

e^{x l n (1 + \frac{9}{x})} \frac{d}{d x} (x ln (1 + \frac{9}{x}))

= e^{x l n (1 + \frac{9}{x})} \frac{d}{d x} (x ln (1 + \frac{9}{x}))

\frac{d}{d x} (x ln (1 + \frac{9}{x})) = ln (1 + \frac{9}{x}) - \frac{9}{x + 9}

= e^{x l n (1 + \frac{9}{x})} (ln (1 + \frac{9}{x}) - \frac{9}{x + 9})

e^{x l n (1 + \frac{9}{x})} = (1 + \frac{9}{x})^{x}

= (ln (1 + \frac{9}{x}) - \frac{9}{x + 9}) (1 + \frac{9}{x})^{x}

What is the derivative of (1+9/x)^x ?
The derivative of (1+9/x)^x is (ln(1+9/x)-9/(x+9))(1+9/x)^x
What is the first derivative of (1+9/x)^x ?
The first derivative of (1+9/x)^x is (ln(1+9/x)-9/(x+9))(1+9/x)^x