What is the ((x+5)e^{1/x})^' ?

The ((x+5)e^{1/x})^' is e^{1/x}-(e^{1/x}(x+5))/(x^2)

What is the first ((x+5)e^{1/x})^' ?

The first ((x+5)e^{1/x})^' is e^{1/x}-(e^{1/x}(x+5))/(x^2)

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((x + 5) e^{\frac{1}{x}})^{^{'}}

e^{\frac{1}{x}} - \frac{e ^{\frac{1}{x}} ( x + 5 )}{x ^{2}}

Solution steps

((x + 5) e^{\frac{1}{x}})^{'}

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x + 5, g = e^{\frac{1}{x}}

= (x + 5)^{^{'}} e^{\frac{1}{x}} + (e^{\frac{1}{x}})^{^{'}} (x + 5)

(x + 5)^{'} = 1

(e^{\frac{1}{x}})^{'} = - \frac{e ^{\frac{1}{x}}}{x ^{2}}

= 1 \cdot e^{\frac{1}{x}} + (- \frac{e ^{\frac{1}{x}}}{x ^{2}}) (x + 5)

Simplify

1 \cdot e^{\frac{1}{x}} + (- \frac{e ^{\frac{1}{x}}}{x ^{2}}) (x + 5) : e^{\frac{1}{x}} - \frac{e ^{\frac{1}{x}} ( x + 5 )}{x ^{2}}

= e^{\frac{1}{x}} - \frac{e ^{\frac{1}{x}} ( x + 5 )}{x ^{2}}

y = 6 x (2 x + 1)^{2} + (2 x + 1)^{3}

y^{^{''}} - 10 y^{^{'}} + 29 y = 0

\frac{d}{d x} (\frac{1}{x} + \frac{3}{x ^{2}} + \frac{2}{x ^{3}})

x \to \infty lim (co d s)

\int_{1}^{e} x (ln (x))^{2} d x

What is the ((x+5)e^{1/x})^' ?
The ((x+5)e^{1/x})^' is e^{1/x}-(e^{1/x}(x+5))/(x^2)
What is the first ((x+5)e^{1/x})^' ?
The first ((x+5)e^{1/x})^' is e^{1/x}-(e^{1/x}(x+5))/(x^2)