Please help with calculus question!

Calculus
Tutor: None Selected Time limit: 1 Day

Determine the equation of the tangent line at x=1 for f(x)=2sqrtx/4x^2-5

Jul 22nd, 2015

Thank you for the opportunity to help you with your question!

is f(x) = 2 sqrt(x/(4x^2-25))?

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 22nd, 2015

after clarification, I will work on it, thanks

Jul 22nd, 2015

No. it's 2sqrtx/4x^2-5

Jul 22nd, 2015

is it sqrt(x)/(4x^2-5) ? or sqrt(x)/(4x^2) -5


Jul 22nd, 2015

2sqrt(x)/(4x^2-5)

Jul 22nd, 2015

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  


Jul 22nd, 2015

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Jul 22nd, 2015
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