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 $d/dx(e^u) = e^u(du)/dx$

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

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

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