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July 26, 2014

July 26, 2014

Posted by **Robin** on Thursday, September 29, 2011 at 9:53pm.

Use this to find the equation of the tangent line to the curve y=(3x)/(1+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 -
**Reiny**, Thursday, September 29, 2011 at 10:21pmI will assume you meant

f(x) = 3x/(1 + x^2)

then by the quotient rule

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

f'(4) = (3(17) - 12(8))/289 = -45/289

so now you have the slope m = -45/289

and your strange point (4, .70588)

Use your grade 9 method of y=mx + b

to find the equation.

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