What is the derivative of 4x^{1/x} ?

The derivative of 4x^{1/x} is 4x^{(1-2x)/x}(-ln(x)+1)

What is the first derivative of 4x^{1/x} ?

The first derivative of 4x^{1/x} is 4x^{(1-2x)/x}(-ln(x)+1)

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

4 x^{\frac{1 - 2 x}{x}} (- ln (x) + 1)

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

= 4 e^{\frac{1}{x} l n (x)} \frac{- ln ( x ) + 1}{x ^{2}}

Simplify

4 e^{\frac{1}{x} l n (x)} \frac{- ln ( x ) + 1}{x ^{2}} : 4 x^{\frac{1 - 2 x}{x}} (- ln (x) + 1)

= 4 x^{\frac{1 - 2 x}{x}} (- ln (x) + 1)

y^{^{''}} + y = 0.75 sin (t) - 0.25 sin (3 t)

\int 2 x + 4 d x

x = 2 y, x + 1 = (y - 1)^{2}

5 - x^{2}, x^{2} - 3

\int_{0}^{4} ∣ 2 x - 6 ∣ d x

What is the derivative of 4x^{1/x} ?
The derivative of 4x^{1/x} is 4x^{(1-2x)/x}(-ln(x)+1)
What is the first derivative of 4x^{1/x} ?
The first derivative of 4x^{1/x} is 4x^{(1-2x)/x}(-ln(x)+1)