What is the (1+xe^{2/x})^' ?

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

What is the first (1+xe^{2/x})^' ?

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

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

e^{\frac{2}{x}} - \frac{2 e ^{\frac{2}{x}}}{x}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

1^{'} = 0

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

= 0 + e^{\frac{2}{x}} - \frac{2 e ^{\frac{2}{x}}}{x}

Simplify

= e^{\frac{2}{x}} - \frac{2 e ^{\frac{2}{x}}}{x}

\frac{d}{d x} (cos (5 x) - 1)

\int \frac{1 - cos ( x )}{1 + sin ( x )} d x

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

y^{^{''}} + 16 y^{^{'}} = 0

\int (sin^{2} (7 t)) d t

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