Find an equation of the tangent line to the curve at the point (1, 1).

ln(xe^(x^2))

first I applied the chain rule to get 1/(xe^(x^2)) *d/dx(xe^(x^2)). this is where I get stuck. which exponent do I start with the powers rule on to find this derivative?

Question

Find an equation of the tangent line to the curve at the point (1, 1).

ln(xe^(x^2))

first I applied the chain rule to get 1/(xe^(x^2)) *d/dx(xe^(x^2)).  this is where I get stuck.  which exponent do I start with the powers rule on to find this derivative?

anonymous · Answer

I'll tell you you're making it harder than you need to.  I'll do it using the chain rule and then I'll show the easy way.

You need to do the product rule on the part you were stuck on.

If y = ln(xe^(x^2)), then

$$dy/dx = 1/xe ^{x^2}[1e ^{x^2}+x*e ^{x^2}2x]$$
$$dy/dx = e ^{x^2}(1 + 2x^2)/xe ^{x^2}=(1 + 2x^2)/x$$

Easy way - use logarithmic transformation:

$$y = \ln(xe ^{x^2})=lnx + lne ^{x^2}= \ln x + x^2$$

$$dy/dx = 1/x + 2x$$

anonymous · Answer

thanks I appreciate your help