What is the derivative of (1+3/x ^x) ?

The derivative of (1+3/x ^x) is (ln(1+3/x)-3/(x+3))(1+3/x)^x

What is the first derivative of (1+3/x ^x) ?

The first derivative of (1+3/x ^x) is (ln(1+3/x)-3/(x+3))(1+3/x)^x

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\frac{d}{d x} ((1 + \frac{3}{x})^{x})

(ln (1 + \frac{3}{x}) - \frac{3}{x + 3}) (1 + \frac{3}{x})^{x}

Solution steps

\frac{d}{d x} ((1 + \frac{3}{x})^{x})

Apply exponent rule:

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

(1 + \frac{3}{x})^{x} = e^{x l n (1 + \frac{3}{x})}

= \frac{d}{d x} (e^{x l n (1 + \frac{3}{x})})

Apply the chain rule:

e^{x l n (1 + \frac{3}{x})} \frac{d}{d x} (x ln (1 + \frac{3}{x}))

= e^{x l n (1 + \frac{3}{x})} \frac{d}{d x} (x ln (1 + \frac{3}{x}))

\frac{d}{d x} (x ln (1 + \frac{3}{x})) = ln (1 + \frac{3}{x}) - \frac{3}{x + 3}

= e^{x l n (1 + \frac{3}{x})} (ln (1 + \frac{3}{x}) - \frac{3}{x + 3})

e^{x l n (1 + \frac{3}{x})} = (1 + \frac{3}{x})^{x}

= (ln (1 + \frac{3}{x}) - \frac{3}{x + 3}) (1 + \frac{3}{x})^{x}

\frac{\partial}{\partial x} (\frac{( 2 x - 5 y )}{2 x + 5 y})

\int_{0}^{π} 6 sin^{4} (x) d x

y = r^{2} - x^{2}, y = 0

\frac{\partial}{\partial x} ((ln (x))^{2})

x, at x = 144

What is the derivative of (1+3/x ^x) ?
The derivative of (1+3/x ^x) is (ln(1+3/x)-3/(x+3))(1+3/x)^x
What is the first derivative of (1+3/x ^x) ?
The first derivative of (1+3/x ^x) is (ln(1+3/x)-3/(x+3))(1+3/x)^x