7.6 The second derivative
Differentiating \(f'(x)\) again gives the second derivative, written \[f''(x)\qquad \text {or}\qquad \frac {d^2y}{dx^2}.\]
Note 7.23. The first derivative measures how fast \(y\) is changing; the second measures how fast that is changing — whether the curve is bending upwards or downwards. Nothing new has to be learnt: differentiate once, then differentiate the answer.
Example 7.24. Find \(\frac {d^2y}{dx^2}\) for (a) \(y=x^4-5x^3+2x\), (b) \(y=e^{3x}\), (c) \(y=\frac {2x-3}{x+2}\).
Solution. (a) Differentiate twice in succession: \[\frac {dy}{dx}=4x^3-15x^2+2,\qquad \frac {d^2y}{dx^2}=12x^2-30x.\]
(b) Each differentiation brings down another factor of \(3\): \[\frac {dy}{dx}=3e^{3x},\qquad \frac {d^2y}{dx^2}=9e^{3x}.\]
(c) By the quotient rule, with \(u=2x-3\) and \(v=x+2\): \[\frac {dy}{dx}=\frac {2(x+2)-(2x-3)(1)}{(x+2)^2}=\frac {2x+4-2x+3}{(x+2)^2} =\frac {7}{(x+2)^2}=7(x+2)^{-2}.\] Writing it as a power makes the second differentiation a chain rule rather than another quotient rule: \[\frac {d^2y}{dx^2}=7\cdot (-2)(x+2)^{-3}\cdot 1=-\frac {14}{(x+2)^3}.\]
Note 7.25. Tidying \(\frac {dy}{dx}\) into index form before differentiating again, as in (c), saves a great deal of work. Applying the quotient rule a second time gives the same answer after much more algebra and many more chances to slip.
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