# What is the equation of the tangent line of #f(x) =1/ (1+2x^2) # at # x = 3#?

First derivation and

Using the first derivation, this is fairly simple.

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To find the equation of the tangent line of f(x) = 1/(1+2x^2) at x = 3, we need to find the derivative of f(x) and evaluate it at x = 3.

The derivative of f(x) is given by f'(x) = -4x/(1+2x^2)^2.

Evaluating f'(x) at x = 3, we get f'(3) = -4(3)/(1+2(3)^2)^2 = -12/(1+18)^2 = -12/19^2.

Using the point-slope form of a line, the equation of the tangent line is y - f(3) = f'(3)(x - 3).

Substituting the values, we have y - f(3) = -12/19^2(x - 3).

Simplifying further, the equation of the tangent line is y = -12/19^2(x - 3) + f(3).

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