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

Question

2219 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-17T16:40:55

Use logarithmic differentiation to obtain the answer:
#dy/dx = ln(5)5^(-1/x)/x^2#

Let #y = 5^(-1/x)#

Use the natural logarithm on both sides:

#ln(y) = ln(5^(-1/x))#

On the right, use the property #ln(a^b) = (b)ln(a)#:

#ln(y) = (-1/x)ln(5) = -ln(5)/x#

Differentiate both sides:

#(d(ln(y)))/dx = (d(-ln(5)/x))/dx#

#1/ydy/dx = ln(5)/x^2#

Multiply both sides by y:

#dy/dx = ln(5)y/x^2#

Substitute #5^(-1/x)# for y:

#dy/dx = ln(5)5^(-1/x)/x^2#