Determine the equation of the tangent line at x=1 for f(x)=2sqrtx/4x^2-5
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is f(x) = 2 sqrt(x/(4x^2-25))?
after clarification, I will work on it, thanks
No. it's 2sqrtx/4x^2-5
is it sqrt(x)/(4x^2-5) ? or sqrt(x)/(4x^2) -5
f'(x) = 1/sqrt(x) (4x^2-5) - 2sqrt(x) *8x/(4x^2-5)^2
f(1) = 2*1/(-1) = -2
f'(1) = -1 -2 *8 / (-1)^2 = -17
the tangent line is
y +2 = -17*(x-1)
y = -17x+15
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