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利用导数求解曲线的切线问题是一种常见的题型,下面通过一例及例题的五个变式对与曲线的切线有关的问题作简单的分析,以供参考,望对你学习这一部分有所帮助.例已知函数f(x)=x~3+x-16,求曲线y=f(x)在点(2,-6)处的切线的方程.解析由题意得,函数f(x)的导数f’(x)=(x~3+x-16)’=3x~2+1,则f’(2)=3×2~2+1=13,即在点(2,-6)处的切线的斜率为k=13,所以切线的方程为:y-(-6)=13(x-2),即y=13x-32.点评求解在某点处的切线方程,关键在于灵活运用x=x_0处的导数就是该点处的切线的斜率这一性质.由导数的几何意义可知,点(x_0,f(x_0))处的切线方程y=f’(x_0)(x-x_0)+f(x_0).
The use of derivatives to solve the problem of tangent curve is a common problem type, the following five examples and a sample of the problem with the tangent of the curve for simple analysis for reference, hope you learn this part of the For example, we know the equation of tangent at the point (2, -6) of the curve y = f (x) with the function f (x) = x ~ 3 + x- (2) = 3 × 2 ~ 2 + 1 = 13, that is, at the point (2, x) 6) = 13 (x-2), that is, y = 13x-32. Comment To solve the tangent equation at a point, The key point is that the flexibility of using the derivative at x = x_0 is the nature of the slope of the tangent at that point. The geometrical significance of the derivative shows that the tangent equation at the point (x_0, f (x_0)) y = f ’(x_0) x-x_0) + f (x_0).