Try $f(x) = \frac{2x+3}{x^2+5}$, eq of tangent line at $x=1$
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f(x) = (2x + 3) / (x^2 + 5) f'(x) = [(2)(x^2 + 5) - (2x + 3)(2x)] / (x^2 + 5)^2 f'(x) = (2x^2 + 10 - 4x^2 - 6x) / (x^2 + 5)^2 f'(x) = (-2x^2 - 6x + 10) / (x^2 + 5)^2 Show more…
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