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- Thread starter noboost4you
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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.

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HallsofIvy

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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

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Originally 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)

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

- #5

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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

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...

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