Find the derivative of f(t) = (2)/(2t-5)^2

Answer 1

f(t)=2 / (2t-5)² we find f'(t) by using the quotient rule: (d/dx) [f(x) / g(x)] = [f'(x)g(x) - g'(x)f(x)] / [g(x)]²

so for f(t), we have f(x)=2, f'(x)=0 and g(x) = (2x-5)², g'(x) = 2(2x-5)(2)

so f'(t)=[0 - 2(2)(2t-5)(2)] /(2t-5)⁴ combine like terms: [-8(2t-5)] /(2t-5)⁴

then noticing the (2t-5) on top will cancel will one of the 4 on bottom to give you

f'(t) = -8 /(2t-3)³