- #1

- 61

- 0

- Thread starter noboost4you
- Start date

- #1

- 61

- 0

- #2

- 515

- 1

After that you know the slope of the tangent and you also know one point of the tangent [that is also (1, e) in your case] and you can easily determine it's equation.

- #3

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 961

The derivative, f', of a function f can be

In order to find the equation of the tangent line to y= f(x) at the point (x

In this problem f(x)= xe^(1/x^2) + ln(3 - 2x^2) so

f'= (1)e^(1/x^2)+ x(1/x^2)(-2/x^3)e^(1/x^2)+ (1/(3-2x^2))(-4x)

(Notice use of product rule and chain rule)

f(1)= e+ ln(1)= e so (1, e)

f'(1)= e- 2e- 4= -e-4

The equation of the tangent line to y= f(x) at (1,e) is

y= (-e-4)(x-1)+ e.

- #4

- 61

- 0

The second part of that derivative is throwing me off. Wouldn't the derivative of xe^(1/x^2) be: (1)e^(1/x^2) + e^(1/x^2)(x)(-2/x^3) ?? I don't know how you incorporated (1/x^2) in that? Please shine some light on that problem. Thanks againOriginally posted by HallsofIvy

In this problem f(x)= xe^(1/x^2) + ln(3 - 2x^2) so

f'= (1)e^(1/x^2)+x(1/x^2)(-2/x^3)e^(1/x^2)+ (1/(3-2x^2))(-4x)

(Notice use of product rule and chain rule)

- #5

- 61

- 0

Well I guess that really wasn't that big of a deal because when f'(1) is plugged in, that extra (1/x^2) equals 1 anyhow. I just didn't see how that came into the derivative...Originally posted by noboost4you

The second part of that derivative is throwing me off. Wouldn't the derivative of xe^(1/x^2) be: (1)e^(1/x^2) + e^(1/x^2)(x)(-2/x^3) ?? I don't know how you incorporated (1/x^2) in that? Please shine some light on that problem. Thanks again

- Last Post

- Replies
- 5

- Views
- 9K

- Last Post

- Replies
- 5

- Views
- 7K

- Replies
- 5

- Views
- 7K

- Replies
- 5

- Views
- 4K

- Replies
- 1

- Views
- 1K

- Replies
- 1

- Views
- 2K

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 980

- Replies
- 5

- Views
- 3K

- Last Post

- Replies
- 2

- Views
- 2K