Question

Help with equations of tangents for exponential functions please?

Answer 1

\(y-e^{xy}=0\implies\frac{dy}{dx}-\frac{d(e^{xy})}{dx}=0\\\frac{dy}{dx}-(e^{xy}()y-\frac{dy}{dx}x)=0\\\frac{dy}{dx}=\frac{e^{xy}y}{1-xe^{xy}}\)
to find the slope at (0,1) we plug in..
\(\frac{dy}{dx}|_{(0,1)}=1\)
using our point again for the line y=mx+b with slope 1
so
\(y=1x+b\implies y=x+1\)

Answer 2

Thanks, I keep getting a slope of 3 for b) and I don't think it's correct ):

Answer 3

\(x^2e^y=1\\2xe^y+x^2\frac{d(e^y)}{dx}=0\\2xe^y+x^2e^y\frac{dy}{dx}=0\\\frac{dy}{dx}=-\frac{2xe^y}{x^2e^y}=-\frac{2}{x}\)
so slope is -2 when x = 1

Answer 4

ohh, I think i found my error. I didn't put the dy/dx when I derived for e^y because I assumed that it's derivative was the same like how we were taught for e^x.

Answer 5

thank you so much! c:

Answer 6

correct, we still use the chain rule on e^x we just dont show it
if y = e^x
then y'=e^x*(d(x)\dx)=e^x*1=e^x

Answer 7

Got it, thanks for clarifying that for me, have a good night!