# Calculus

posted by
**Anonymous**
.

f(x)=(5x)/(1+2x^2).

f'(4)=-75/289

f'(x)=(5-5x^2)/1+2x^2+x^4)

Use this equation of the tangent line to the curve y=5x/(1+x^2) at the point (4, 1.17647). The equation of the tangent line can be written in the form y=mx+b. m=-75/289. What is b?