Second-Order Derivatives

MATSecond-Order Derivatives

The second-order derivative $\dfrac{d^2y}{dx^2}$ is the derivative of $\dfrac{dy}{dx}$ and measures curvature/concavity. A signature ISC theme is proving a differential equation that $y$ satisfies, by computing $y'$ and $y''$ and eliminating the constants.

These proofs often start from $y=e^{a\sin^{-1}x}$ or $y=(\sin^{-1}x)^2$ and rely on squaring relations to remove surds.

Definition

\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\!\left(\dfrac{dy}{dx}\right)

y''=\dfrac{d^2y}{dx^2}

; the derivative of the first derivative w.r.t. the same variable

x

Harmonic relation

y = A\sin x + B\cos x \;\Rightarrow\; \dfrac{d^2y}{dx^2} + y = 0

A,B

arbitrary constants; differentiate twice and add to eliminate them.

Result for

y=(\sin^{-1}x)^2

(1-x^2)\,y'' - x\,y' - 2 = 0

Start from

(1-x^2)(y')^2=4y

, differentiate once more and cancel

2y'

Result for

y=e^{a\sin^{-1}x}

(1-x^2)\,y'' - x\,y' - a^2 y = 0

\sqrt{1-x^2}\,y'=a y

; square to remove the surd, then differentiate and simplify.

Compute $y'$ first, then differentiate again (using product/quotient/chain rules) to get $y''$ .
To prove a differential equation, find $y'$ , often square or rearrange to eliminate surds/inverse-trig, then differentiate again.
For $y=e^{m\tan^{-1}x}$ : $(1+x^2)y'=my$ , differentiating gives $(1+x^2)y''+(2x-m)y'=0$ .
After differentiating a squared relation like $(1-x^2)(y')^2=ky$ , divide through by the common factor $2y'$ (valid where $y'\ne0$ ).
Keep arbitrary constants until they cancel; the final differential equation should be free of them.
Watch the sign of the $x y'$ term: it is $-x y'$ for the $\sin^{-1}$ -based functions above and $+(2x-m)y'$ for the $e^{m\tan^{-1}x}$ case.
Verify the order matches: an $n$ -constant family yields an $n$ -th order differential equation.

Where the marks go

Forgetting the chain rule on the second differentiation, especially with $\sqrt{1-x^2}$ or inverse-trig terms.
Sign errors in the $x y'$ term when eliminating constants.
Failing to square the first-derivative relation to clear the surd before differentiating again.
Leaving the arbitrary constants ( $A$ , $B$ , $a$ , $m$ ) in the final result instead of eliminating them.

How the board asks it

Derive / proveeliminating constants from $y'$ and $y''$
If $y = e^{a\sin^{-1}x}$ , prove that $(1-x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} - a^2 y = 0$ .
Numericalproduct and chain rule
If $y = x^3 \log x$ , find $\dfrac{d^2y}{dx^2}$ and evaluate it at $x = 1$ .
Numericalparametric differentiation
If $x = a(\theta + \sin\theta)$ and $y = a(1 - \cos\theta)$ , find $\dfrac{d^2y}{dx^2}$ at $\theta = \dfrac{\pi}{2}$ .
Derive / prove $e^{m\tan^{-1}x}$ relation $(1+x^2)y'=my$
If $y = e^{m\tan^{-1}x}$ , prove that $(1+x^2)\dfrac{d^2y}{dx^2} + (2x - m)\dfrac{dy}{dx} = 0$ .
Derive / prove $\cos(\log x)$ and $\sin(\log x)$ family
If $y = A\cos(\log x) + B\sin(\log x)$ , show that $x^2\dfrac{d^2y}{dx^2} + x\dfrac{dy}{dx} + y = 0$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.