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

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Answer 1 · 2017-01-17T16:40:55

Use logarithmic differentiation to obtain the answer:
$\frac{d y}{d x} = ln (5) \frac{5^{- \frac{1}{x}}}{x^{2}}$ $dy/dx = ln(5)5^(-1/x)/x^2$

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

Use the natural logarithm on both sides:

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

On the right, use the property $ln (a^{b}) = (b) ln (a)$ $ln(a^b) = (b)ln(a)$ :

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

Differentiate both sides:

$\frac{d (ln (y))}{d x} = \frac{d (- \frac{ln (5)}{x})}{d x}$ $(d(ln(y)))/dx = (d(-ln(5)/x))/dx$

$\frac{1}{y} \frac{d y}{d x} = \frac{ln (5)}{x^{2}}$ $1/ydy/dx = ln(5)/x^2$

Multiply both sides by y:

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

Substitute $5^{- \frac{1}{x}}$ $5^(-1/x)$ for y:

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

What is the derivative of 5^(-1/x)5−1x5^(-1/x)?