Find the linearization L(x) of y=e^(3x)ln(x) at a=1

Question

anonymous · Answer

derivative of the function: e^(3x)(3xlog(x)+1)/x

helder_edwin · Answer

do u mean
$$ y-y(1)=y'(1)(x-1) $$
?

anonymous · Answer

Yea I think thats what it mean, I think linerization is just another word for finding equation of the tangent line but I am not 100% sure

helder_edwin · Answer

u can aproximate any curve with its derivatives. the easiest would be the tangent line, as you said

anonymous · Answer

Yea what I did so far was use the equation you used, however when I typed in numbers I got them wrong so maybe I am solving it wrong?

helder_edwin · Answer

since
$$ y=e^{3x}\ln x $$
then
$$ y(1)=e^3\ln 1=0 $$
right?

helder_edwin · Answer

then
$$ y'=3e^{3x}\ln x+e^{3x}\frac{1}{x}=e^{3x}\left(\ln x+\frac{1}{x}\right) $$
then
y'(1)=e^3(\ln 1+1)=e^3

anonymous · Answer

Yea, and then I got 50 for the answer of the derivative form one.

helder_edwin · Answer

$$ y'(1)=e^3(\ln 1+1)=e^3 $$

helder_edwin · Answer

so the tangent line would be
$$\large y=e^3(x-1) $$

anonymous · Answer

ah I got it, for some reason I kept getting 50 for the one answer l(x)=20.08553692x-20.08553692
Thanks you!

helder_edwin · Answer

u r welcome