What is the derivative of $y = 5^{- \frac{1}{x}}$ $y=5^(-1/x)$ ?

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Answer 1 · 2016-08-13T06:53:35

$\frac{d y}{d x} = \frac{ln 5}{x^{2} \cdot 5^{\frac{1}{x}}}$ $dy/dx = ln5/(x^2*5^(1/x))$

$y = 5^{- \frac{1}{x}}$ $y=5^(-1/x)$

$ln y = - \frac{1}{x} \cdot ln 5 = - ln 5 \cdot x^{- 1}$ $lny =-1/x*ln5 = -ln5 *x^(-1)$

$\frac{1}{y} \frac{d y}{d x} = ln 5 \cdot x^{- 2}$ $1/y dy/dx = ln5*x^(-2)$
(Implicit differentiation, standard differential and power rule)

$\frac{d y}{d x} = ln 5 \cdot x^{- 2} \cdot y$ $dy/dx = ln5*x^(-2) * y$

$= ln 5 \cdot x^{- 2} \cdot 5^{- \frac{1}{x}}$ $=ln5*x^(-2) * 5^(-1/x)$

$= \frac{ln 5}{x^{2} \cdot 5^{\frac{1}{x}}}$ $=ln5/(x^2*5^(1/x))$

What is the derivative of y=5^(-1/x)y=5−1xy=5^(-1/x)?