What is the first (x+4/x)^' ?

The first (x+4/x)^' is 1-4/(x^2)

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(x+4/x)^'

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

1 - \frac{4}{x ^{2}}

Solution steps

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

Apply the Sum/Difference Rule:

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

= x^{^{'}} + (\frac{4}{x})^{^{'}}

x^{'} = 1

(\frac{4}{x})^{'} = - \frac{4}{x ^{2}}

= 1 - \frac{4}{x ^{2}}

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

\int (x + 2)^{5} d x

\int \frac{1}{3} e^{- \frac{x}{3}} d x

x \to - 4 - lim (x^{2} + \frac{2}{x + 4})

\frac{d}{d x} (\frac{2}{x} - 3 x)