Question

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

Answer 1

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`

Answer 2

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`