What is the derivative of #y = ln(xe^(x^2))# at the point (1,1)?

2 Answers
Oct 20, 2017

y = 3(x-1) + 1

Explanation:

STEP 1: Identify the rules needed to derive the function. For this equation, both the Chain Rule and Product Rule will be needed.
(Just a reminder, the Chain Rule: #(d/dx)# f(g(x)) = #f'(g(x))*g'(x)# and the Product Rule: #(d/dx) f(x)*g(x) = f'(x)*g(x) + f(x)*g'(x)#)

Thus, using these two rules, you should get #y' = (1/(xe^(x^2))) * ((e^(x^2)) + (x)(2x*e^(x^2)))#

STEP 2: Simplify the derivative

#y' = (e^(x^2) + (2x^2)(e^(x^2)))/(xe^(x^2))#

STEP 3: You can now plug in the x = 1

y'(1) = #(e^(1^2) + (2(1)^2)(e^(1^2)))/((1)e^(1^2))#

y'(1) = #(e + 2e)/e#

y'(1) = #(3e)/e#

y'(1) = 3

Therefore, at the point (1,1), m = 3

STEP 4: Plug in the slope and points into the Point Slope Form Equation

#y - 1 = 3(x-1)#

#y = 3(x-1) + 1#

Oct 20, 2017

The answer is 3.

Explanation:

First, recall that #ln(ab)=lna+lnb#
From the question, we can see that #y=lnx+ln(e^(x^2))#

Then, the form of #e^(x^2)# still looks like complicated to solve with. However, using the rule of #lne^k=klne#, we can further simplify it and
#y=lnx+(x^2)lne#

Furthermore, we know that #lne=1#
So, we can simply know that #y=lnx+x^2#

And we can start differentiate it by using #d(lnx)/dx=1/x# and power rule.

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

At the point #(1,1)# ,

#dy/dx=1/(1)+2(1)=1+2=3#