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.

Fig. 19.	Fig. 20.

(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