How do you differentiate #y=x^(8x)#?

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Answer 1 · 2016-04-18T13:37:58

Use logarithmic differentiation (or its equivalent exponential form).

#y=x^(8x)#

Logarithmic Differentiation

Take the logarithm of both sides.

#lny=ln(x^(8x))=8xlnx#. Now differentiate implicitly.

#1/y dy/dx = 8lnx+8x(1/x) = 8lnx+8#

#dy/dx = y(8lnx+8) = x^(8x)(8lnx+8)#.

Exponential Equivalent

#y=x^(8x) = e^(ln(x^(8x))) = e^(8xlnx)#

Differentiate using #d/dx(e^u) = e^u(du)/dx#

#dy/dx = e^(8xlnx)(8lnx+8)#

# = x^(8x)(8lnx+8)#