How do you find the derivative of $y=2^(-3/x)$ ?

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Answer 1 · 2017-08-13T18:18:08

$dy/dx = 2^(-3/x)(3/x^2)ln2$

Use logarithmic differentiation.

$lny = ln(2^(-3/x))$

$lny = -3/xln(2)$

$lny = -3x^-1(ln2)$

$lny = -3ln2x^-1$

$1/y(dy/dx)= 3ln2x^(-2)$

$dy/dx = 2^(-3/x)(3/x^2)ln2$

Hopefully this helps!

How do you find the derivative of y=2^(-3/x)y=2^(-3/x)?