the normal line of y = ((3x-1)^2)((x-1)^3) at x = 0

Question

campbell_st · Answer

so you need to find the derivative, which will require the product and chain rules for differentiation. 


but before differentiating, you need a point on the curve... so substitute x = 0 to find a y value, this will be the point you use to find the equation of the normal. 

the derivative gives the equation of the slope of a tangent at any point on the curve. so find the derivative

Then substitute x = 0 into the equation to find the slope of the tangent. 

then the slope of the normal is the negative reciprocal of the tangent slope

the you will have a point and a slope, so you can find the equation of the normal

irishboy123 · Answer

$$ y = (3x-1)^2(x-1)^3 $$

i'd use the product rule....hopefully setting x=0 will simplify some stuff.

freckles · Answer

I would use log differentiation because I'm weird like that

irishboy123 · Answer

that's classy!!

$\ln y = 2 \ln (3x-1) + 3 \ln (x-1)$

$\frac{1}{y} y' = .....$