What is (\partial}{\partial x}(B(\frac{G(B,x))/a)^{1/x}) ?

The answer to (\partial}{\partial x}(B(\frac{G(B,x))/a)^{1/x}) is

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(\partial}{\partial x}(B(\frac{G(B,x))/a)^{1/x})

\frac{\partial}{\partial x} (B (\frac{G ( B , x )}{a})^{\frac{1}{x}})

B e^{\frac{1}{x} l n (\frac{G ( B , x )}{a})} ((- \frac{1}{x ^{2}}) ln (\frac{G ( B , x )}{a}) + \frac{1}{\frac{G ( B , x )}{a}} \cdot \frac{1}{a} \frac{\partial}{\partial x} (G (B, x)) \frac{1}{x})

Solution steps

\frac{\partial}{\partial x} (B (\frac{G ( B , x )}{a})^{\frac{1}{x}})

Treat

a, B

as a constant

Take the constant out:

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

= B \frac{\partial}{\partial x} ((\frac{G ( B , x )}{a})^{\frac{1}{x}})

Apply exponent rule:

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

(\frac{G ( B , x )}{a})^{\frac{1}{x}} = e^{\frac{1}{x} l n (\frac{G ( B , x )}{a})}

= B \frac{\partial}{\partial x} (e^{\frac{1}{x} l n (\frac{G ( B , x )}{a})})

Apply the chain rule:

e^{\frac{1}{x} l n (\frac{G ( B , x )}{a})} \frac{\partial}{\partial x} (\frac{1}{x} ln (\frac{G ( B , x )}{a}))

= e^{\frac{1}{x} l n (\frac{G ( B , x )}{a})} \frac{\partial}{\partial x} (\frac{1}{x} ln (\frac{G ( B , x )}{a}))

\frac{\partial}{\partial x} (\frac{1}{x} ln (\frac{G ( B , x )}{a})) = (- \frac{1}{x ^{2}}) ln (\frac{G ( B , x )}{a}) + \frac{1}{\frac{G ( B , x )}{a}} \cdot \frac{1}{a} \frac{\partial}{\partial x} (G (B, x)) \frac{1}{x}

= B e^{\frac{1}{x} l n (\frac{G ( B , x )}{a})} ((- \frac{1}{x ^{2}}) ln (\frac{G ( B , x )}{a}) + \frac{1}{\frac{G ( B , x )}{a}} \cdot \frac{1}{a} \frac{\partial}{\partial x} (G (B, x)) \frac{1}{x})

\frac{\partial}{\partial x} (- 2 x^{3} + sin (x y) + 4 e^{4 y})

x \to 0 lim (∣ x ∣ ln ∣ x ∣)

\frac{1}{s ^{2}} (1 - e^{- s})

\frac{1}{( x ^{2} + 8 ) ^{5}}

What is (\partial}{\partial x}(B(\frac{G(B,x))/a)^{1/x}) ?
The answer to (\partial}{\partial x}(B(\frac{G(B,x))/a)^{1/x}) is