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