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November 28, 2014

November 28, 2014

Posted by **Allison** on Thursday, September 29, 2011 at 1:07pm.

find f′(4).

Use this to find the equation of the tangent line to the curve y=3x1+x2 at the point (4,0.70588). The equation of this tangent line can be written in the form y=mx+b where m is:

and where b is:

- Calculus -
**Steve**, Thursday, September 29, 2011 at 2:46pmf(x) = 3x/(1+x²)

f'(x) = [(3)(1+x²) - (3x)(2x)]/(1+x²)²

= (3+3x²-6x²)/(1+x²)² = (3-3x²)/(1+x²)²

f(4) = 12/17 = 0.70588

f'(4) = -9/289 = -0.03114

So, we want a line through (4/0.70588) with slope -0.03114

(y-0.70588)/(x-4) = -0.03114

y = -0.3114x + 0.83044

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