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Well, dy/dx is the gradient at a specific point of a curve.

ok, so i need to find the equation of the tangent line

im meant y being negative...

Got it. So i solved it and got: \[y(x) = x^2 +C\] What do you think?

yep well close, i get constant in front
y = Cx^2

ahhhhh, yes. Because 2lnx + C becomes the exponent for e and e^C becomes C

haha yeah was just going to say that :)

I always mess up whenever e or ln are involved

Thank you for your help, you got me out of a pickle :)

no problem