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# The Second Derivative

by ongzz(3 minute read · 0 views)

# the derivative

when i first learned calculus, i was taught how to take functions' derivatives, and to set them equal to zero to find their turning points. i was BLOWN away by this, thinking to myself why it wasn't taught earlier on.

looking back, i think the reason i felt like this was because we were always asked to find the turning point of a parabola, something like $f(x) = 3x^2-12x+16$ by completing the square and writing it in vertex form (which looks like $f(x)=a(x-h)^2+k$).

i didn't really like it - at the time, we were taught how to "completing the square", butweren't told the "why" - i remember feeling super puzzled about why we had to divide by 2 before putting it into parentheses, so each time i did it i felt even more weirded out, like there's some hidden truth (eventually i figured it out though - just expand compare the coefficients of $(x-h)^2+k$ and $x^2+bx+c$ and solve for $h$)

going back to derivatives - the process felt magical. even with only the power rule, i could now solve for turning points very easily, but setting the derivative equal to $0$.

play around with it here! (remember - derivative of $f$ at $x$ $=$ slope of the tangent line at $x$, and slope = 0 at turning point)

but derivatives are not what this blog is about lmao

# the second derivative test

shortly after learning that, we were taught second derivatives, taking the derivative of a derivative, derivative-ception. putting it in practice, it would look like

\begin{aligned} f(x) &= 3x^2 - 12x +16 \\ f'(x) &= 6x - 12 \\ f''(x) &= 6 \end{aligned}

apparently by doing this, we could know whether or not the function is minimum or maximum just by doing math, without drawing graphs whatsoever!

determining a function's nature still felt a bit off, though - it didn't click for me instantly back then.

we were asked to remember that if $f''(x)>0$, the point would be a minimum; and if $f''(x)<0$, the point would be a maximum. isn't it weird though? maximum point implying a negative second derivative??

the way most people explain it is to think of the slope around that turning point - for a maximum turning point, the slope keeps decreasing, from positive slope, to 0 slope, to negative slope.

the same logic could be applied to a minimum point. so, because the 2nd derivative is the rate of change of the derivative, it would be negative at the maximum and positive at the minimum.

# another perspective

the way i got around memorising this was thinking of it as a "force" - positive $f''(x)$ means there's a force "pulling" the graph upwards, and negative $f''(x)$ means there's a force "pulling" the graph downwards.

i'm not sure if this is a correct justification, but because $F=ma$ and $a=\frac{d^2}{dt^2}s$, the rate of change of the rate of change of displacement is acceleration, which is just force scaled by $\frac{1}{m}$. ($a=\frac{F}{m}$)

in our case, the 2nd derivative would be analagous to $F$orce and the $x$-axis as time, with the graph representing displacement over time.

# conclusion

• maximum: $f'(x) = 0, f''(x) < 0$
• minimum: $f'(x) = 0, f''(x) > 0$
• how to memorise? forces!