Question

differentiate y= (e^-2x) ln(x^2)

Answer 1

is it
\[y=e^{-2x}\ln(x^2)\]?

Answer 2

yupp haha

Answer 3

Sooo product rule, yes? :)
\[\Large y'\quad=\quad \color{royalblue}{\left[e^{-2x}\right]'}\ln(x^2)+e^{-2x}\color{royalblue}{\left[\ln(x^2)\right]'}\]

Answer 4

ohh okay that clears things up a bit:) haha
but I don't know how to find the first derivatives of the two

Answer 5

Hmm, remember this derivative?\[\Large (e^x)'\quad=\quad ?\]

Answer 6

e^x so does e^-2x = -2e^-2x and ln(x^2) become 1/(x^2)?

Answer 7

Hmm your first one looks good.\[\Large y'\quad=\quad \color{orangered}{\left[-2e^{-2x}\right]}\ln(x^2)+e^{-2x}\color{royalblue}{\left[\ln(x^2)\right]'}\]

Answer 8

\[\Large \left[\ln(x^2)\right]'\quad=\quad \frac{1}{x^2}(x^2)' \quad=\quad \frac{2}{x}\]

Answer 9

You could also use rules of exponents to do that if you needed :)\[\Large \left[\ln(x^2)\right]'\quad=\quad\left[2\ln(x)\right]'\quad=\quad2\left[\ln(x)\right]'\quad=\quad 2\frac{1}{x}\]