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

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/07 02:08:36
1道求导题y=5000/(x^2-2x+10)^3Show that y'=-3 at the point(2,5)

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

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
y'(2)=-30000/(2^2-2*2+10)^4=-30000/10^4=-3

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