Find the linearization L(x) of y=e8xln(x) at a=1.

Answer 1

Let's denote f(x) = y. The linearization L(x) of y at a point x= a is given as:

L(x) = f(a) + f^'(a) (x-a).

For y = f(x) = e^8x ln(x), we have:

f(1) = e⁸ ln(1) = 0.

Also, using the chain rule for derivatives: f^'(x) = 8 e^8x ln(x) + (e^8x)/x , and for a=1 we obtain: f^'(1)= e⁸.

Subsequently L(x) = f(1) + f^'(1) (x-1) = e⁸ (x-1).