Page:Calculus Made Easy.pdf/104

 ascends at $$45^\circ$$; for whatever values we give to $$x$$ to the right, we have an equal $$y$$ to ascend. The line has a gradient of $$1$$ in $$1$$.

Now differentiate $$y=x+b$$, by the rules we have already learned (pp. 22 and 26 ante), and we get $$\dfrac{dy}{dx}=1$$.

The slope of the line is such that for every little step $$dx$$ to the right, we go an equal little step $$dy$$ upward. And this slope is constant–always the same slope.

(2) Take another case: $y=ax+b$. We know that this curve, like the preceding one, will start from $$a$$ height $$b$$ on the $$y$$-axis. But before we draw the curve, let us find its slope by differentiating; which gives us $$\dfrac{dy}{dx}=a$$. The slope will be constant, at an angle, the tangent of which is here called $$a$$. Let us assign to $$a$$ some numerical value–say $$\tfrac{1}{3}$$. Then we must give it such a slope that it ascends $$1$$ in $$3$$; or