Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/56



But when we let $$\Delta x \dot= 0$$, the point $$Q$$ will move along the curve and approach nearer and nearer to $$P$$, the secant will turn about $$P$$ and approach the tangent as a limiting position, and we have also

Hence from (B) and (C), $$\tfrac{dy}{dx} =$$ slope of the tangent line $$PT$$. Therefore

The value of the derivative at any point of a curve is equal to the slope of the line drawn tangent to the curve at that point.

It was this tangent problem that led Leibnitz to the discovery of the Differential Calculus.

Find the slopes of the tangents to the parabola $$y = x^2$$ at the vertex, and at the point where $$x = \tfrac{1}{2}$$.


 * Solution. Differentiating by General Rule, §31, we get


 * {| style="width:100%"


 * style="width:3%"|
 * align="center"| $$\frac{dy}{dx} = 2x = $$ slope of tangent line at any point on curve.
 * }


 * To find slope of tangent at vertex, substitute $$x = 0$$ in (A), giving

$$\frac{dy}{dx} = 0$$.


 * Therefore the tangent at vertex has the slope zero; that is, it is parallel to the axis of x and in this case coincides with it.


 * To find slope of tangent at the point $$P$$, where $$x = \tfrac{1}{2}$$, substitute in (A), giving

$$\frac{dy}{dx} = 1$$;


 * that is, the tangent at the point $$P$$ makes an angle of 45° with the axis of $$x$$.