1道求导题y=5000/（x^2-2x+10)^3Show that y'=-3 at the point(2,5)

答：
y=5000/(x²-2x+10)³
=5000×(x²-2x+10)^(-3)
y'(x)=5000×(-3)×[(x²-2x+10)^(-3-1)]×(2x-2)
y'(x)=-30000×(x-1) /(x²-2x+10)^4
在点（2,5）处：x=2
y(2)=5000/(2²-2×2+10)³
=5000/1000
=5
所以：点(2,5)在曲线上
所以：
y'(2)=-30000×(2-1)/(2²-2×2+10)^4
=-30000/10000
=-3
所以：y'=-3 at the point(2,5)

求导：已知y=5000/(x²-2x+10)³
y=5000(x²-2x+10)⁻³
y'=-15000(2x-2)(x²-2x+10)⁻⁴=-30000(x-1)/(x²-2x+10)⁴

求导：已知y=5000/(x²-2x+10)³
y=5000(x²-2x+10)⁻³
y'=-15000(2x-2)(x²-2x+10)⁻⁴=-30000(x-1)/(x²-2x+10)⁴
y'(2)=-30000/(2^2-2*2+10)^4=-30000/10^4=-3

y＝5000/(x²-2x+10)
→y'＝5000·(-3)(x²-2x+10)^(-3-1)·(x²-2x+10)'
→y'＝-30000(x-1)/(x²-2x+10)^4.