What is the derivative of (1+4/x)^x ?

The derivative of (1+4/x)^x is (ln(1+4/x)-4/(x+4))(1+4/x)^x

What is the first derivative of (1+4/x)^x ?

The first derivative of (1+4/x)^x is (ln(1+4/x)-4/(x+4))(1+4/x)^x

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derivative of

(1 + \frac{4}{x})^{x}

(ln (1 + \frac{4}{x}) - \frac{4}{x + 4}) (1 + \frac{4}{x})^{x}

Solution steps

\frac{d}{d x} ((1 + \frac{4}{x})^{x})

Apply exponent rule:

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

(1 + \frac{4}{x})^{x} = e^{x l n (1 + \frac{4}{x})}

= \frac{d}{d x} (e^{x l n (1 + \frac{4}{x})})

Apply the chain rule:

e^{x l n (1 + \frac{4}{x})} \frac{d}{d x} (x ln (1 + \frac{4}{x}))

= e^{x l n (1 + \frac{4}{x})} \frac{d}{d x} (x ln (1 + \frac{4}{x}))

\frac{d}{d x} (x ln (1 + \frac{4}{x})) = ln (1 + \frac{4}{x}) - \frac{4}{x + 4}

= e^{x l n (1 + \frac{4}{x})} (ln (1 + \frac{4}{x}) - \frac{4}{x + 4})

e^{x l n (1 + \frac{4}{x})} = (1 + \frac{4}{x})^{x}

= (ln (1 + \frac{4}{x}) - \frac{4}{x + 4}) (1 + \frac{4}{x})^{x}

\frac{\partial}{\partial x} (2 e^{x - 2} - y)

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

\int sin (20 x) sin (21 x) d x

\int x (\frac{1}{x}) d x

\frac{d}{d x} (\frac{4 x ^{2} + 5}{x ^{2} + 7})

What is the derivative of (1+4/x)^x ?
The derivative of (1+4/x)^x is (ln(1+4/x)-4/(x+4))(1+4/x)^x
What is the first derivative of (1+4/x)^x ?
The first derivative of (1+4/x)^x is (ln(1+4/x)-4/(x+4))(1+4/x)^x