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Parametric and second derivatives

In some problems xx and yy are both given as functions of a third variable tt , typically time or angle , rather than yy being given as a function of xx. Examples: a particle moving along a curve, or a circle parametrised as (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta). The parametric derivative formula expresses dydx\dfrac{dy}{dx} without first eliminating the parameter.

A second derivative d2ydx2\dfrac{d^2 y}{dx^2} is the derivative of dydx\dfrac{dy}{dx}, taken with respect to xx again. Higher orders follow the same pattern.

Parametric derivative

If x=f(t)x = f(t) and y=g(t)y = g(t) are differentiable functions of tt, with f′(t)≠0f'(t) \neq 0, then

dydx=dy/dtdx/dt=g′(t)f′(t).\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}.

This follows from the chain rule: dydt=dydx⋅dxdt\dfrac{dy}{dt} = \dfrac{dy}{dx}\cdot \dfrac{dx}{dt}.

Worked example

For the cycloid x=a(t−sin⁡t),y=a(1−cos⁡t)x = a(t - \sin t), y = a(1 - \cos t):

dxdt=a(1−cos⁡t)\dfrac{dx}{dt} = a(1 - \cos t), dydt=asin⁡t\dfrac{dy}{dt} = a \sin t.

dydx=asin⁡ta(1−cos⁡t)=sin⁡t1−cos⁡t=2sin⁡(t/2)cos⁡(t/2)2sin⁡2(t/2)=cot⁡(t/2)\dfrac{dy}{dx} = \dfrac{a \sin t}{a(1 - \cos t)} = \dfrac{\sin t}{1 - \cos t} = \dfrac{2 \sin(t/2)\cos(t/2)}{2 \sin^2(t/2)} = \cot(t/2).

Second derivative in parametric form

d2ydx2=ddx(dydx)=ddt(dydx)⋅dtdx=(dy/dx)t′dx/dt\dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) = \dfrac{d}{dt}\left(\dfrac{dy}{dx}\right) \cdot \dfrac{dt}{dx} = \dfrac{(dy/dx)'_t}{dx/dt}.

In words: differentiate dydx\dfrac{dy}{dx} as a function of tt, then divide by dxdt\dfrac{dx}{dt}.

Higher-order derivatives

f′′(x)f''(x), f′′′(x)f'''(x), f(n)(x)f^{(n)}(x) , repeated differentiation. Standard formulas:

  • dndxn(xk)=k(k−1)⋯(k−n+1)xk−n\dfrac{d^n}{dx^n}(x^k) = k(k-1)\cdots(k - n + 1) x^{k - n} for k≥nk \ge n.
  • dndxn(eax)=aneax\dfrac{d^n}{dx^n}(e^{ax}) = a^n e^{ax}.
  • dndxn(sin⁡x)=sin⁡(x+nπ/2)\dfrac{d^n}{dx^n}(\sin x) = \sin(x + n\pi/2).
  • dndxn(cos⁡x)=cos⁡(x+nπ/2)\dfrac{d^n}{dx^n}(\cos x) = \cos(x + n\pi/2).
  • dndxn(ln⁡x)=(−1)n−1(n−1)!xn\dfrac{d^n}{dx^n}(\ln x) = \dfrac{(-1)^{n-1}(n-1)!}{x^n}.

Worked examples

Example 1. If x=acos⁡θ,y=asin⁡θx = a\cos\theta, y = a\sin\theta, find dydx\dfrac{dy}{dx} and d2ydx2\dfrac{d^2 y}{dx^2}.

dxdθ=−asin⁡θ\dfrac{dx}{d\theta} = -a\sin\theta, dydθ=acos⁡θ\dfrac{dy}{d\theta} = a\cos\theta. So dydx=−cot⁡θ\dfrac{dy}{dx} = -\cot\theta.

ddθ(−cot⁡θ)=csc⁡2θ\dfrac{d}{d\theta}(-\cot\theta) = \csc^2\theta. Divide by dxdθ=−asin⁡θ\dfrac{dx}{d\theta} = -a\sin\theta: d2ydx2=−csc⁡2θasin⁡θ=−1asin⁡3θ\dfrac{d^2 y}{dx^2} = -\dfrac{\csc^2\theta}{a\sin\theta} = -\dfrac{1}{a \sin^3\theta}.

Example 2. If x=t2,y=t3x = t^2, y = t^3, find dydx\dfrac{dy}{dx} and d2ydx2\dfrac{d^2 y}{dx^2}.

dxdt=2t\dfrac{dx}{dt} = 2t, dydt=3t2\dfrac{dy}{dt} = 3t^2. So dydx=3t2\dfrac{dy}{dx} = \dfrac{3t}{2}.

ddt(3t/2)=3/2\dfrac{d}{dt}(3t/2) = 3/2. Divide by 2t2t: d2ydx2=34t\dfrac{d^2 y}{dx^2} = \dfrac{3}{4t}.

Example 3. If y=asin⁡x+bcos⁡xy = a\sin x + b\cos x, show y′′+y=0y'' + y = 0.

y′=acos⁡x−bsin⁡xy' = a\cos x - b\sin x. y′′=−asin⁡x−bcos⁡x=−yy'' = -a\sin x - b\cos x = -y. So y′′+y=0y'' + y = 0.

Example 4. Find the nn-th derivative of f(x)=sin⁡xf(x) = \sin x.

f(n)(x)=sin⁡(x+nπ/2)f^{(n)}(x) = \sin(x + n\pi/2).

Example 5. If y=easin⁡−1xy = e^{a \sin^{-1} x}, show (1−x2)y′′−xy′−a2y=0(1 - x^2) y'' - x y' - a^2 y = 0.

y′=easin⁡−1x⋅a1−x2=ay1−x2y' = e^{a \sin^{-1} x} \cdot \dfrac{a}{\sqrt{1 - x^2}} = \dfrac{a y}{\sqrt{1 - x^2}}. So y′1−x2=ayy' \sqrt{1 - x^2} = ay, square: (1−x2)(y′)2=a2y2(1 - x^2)(y')^2 = a^2 y^2.

Differentiate: −2x(y′)2+2(1−x2)y′y′′=2a2yy′-2 x (y')^2 + 2(1 - x^2) y' y'' = 2 a^2 y y'. Divide by 2y′2 y' (assuming nonzero): −xy′+(1−x2)y′′=a2y-x y' + (1 - x^2) y'' = a^2 y, i.e., (1−x2)y′′−xy′−a2y=0(1 - x^2) y'' - x y' - a^2 y = 0. ■\blacksquare

Example 6. Find d2ydx2\dfrac{d^2 y}{dx^2} for y=xlog⁡xy = x \log x.

y′=log⁡x+1y' = \log x + 1. y′′=1/xy'' = 1/x.

Try it yourself

  1. If x=at2,y=2atx = at^2, y = 2at, find dydx\dfrac{dy}{dx}.
  2. If x=3cos⁡t,y=4sin⁡tx = 3 \cos t, y = 4 \sin t, find dydx\dfrac{dy}{dx} and d2ydx2\dfrac{d^2 y}{dx^2}.
  3. Find the nn-th derivative of cos⁡x\cos x.
  4. Find the nn-th derivative of eaxe^{ax}.
  5. If y=tan⁡−1xy = \tan^{-1} x, find y′′y''.
  6. If y=acos⁡(log⁡x)+bsin⁡(log⁡x)y = a\cos(\log x) + b\sin(\log x), show x2y′′+xy′+y=0x^2 y'' + x y' + y = 0.
  7. If y=sin⁡−1xy = \sin^{-1} x, find y′′y'' and show (1−x2)y′′−xy′=0(1 - x^2) y'' - x y' = 0.
  8. Find d2ydx2\dfrac{d^2 y}{dx^2} for y=ln⁡(sin⁡x)y = \ln(\sin x).
  9. If x=a(θ−sin⁡θ),y=a(1−cos⁡θ)x = a(\theta - \sin\theta), y = a(1 - \cos\theta) (the cycloid), compute d2ydx2\dfrac{d^2 y}{dx^2} at θ=π/2\theta = \pi/2.
  10. If y=excos⁡xy = e^x \cos x, show y′′−2y′+2y=0y'' - 2 y' + 2 y = 0.
  11. Find the 55-th derivative of ln⁡x\ln x.
  12. If x=sec⁡θ,y=tan⁡θx = \sec\theta, y = \tan\theta, find dydx\dfrac{dy}{dx}.
  13. If y=xxy = x^x, find y′′y'' at x=1x = 1.
  14. Find d2ydx2\dfrac{d^2 y}{dx^2} for x=acos⁡3t,y=asin⁡3tx = a \cos^3 t, y = a \sin^3 t (the astroid).

Pitfalls / Tricks

  • For parametric d2ydx2\dfrac{d^2 y}{dx^2}, don't divide the second derivatives directly. Differentiate dydx\dfrac{dy}{dx} with respect to tt first, then divide by dxdt\dfrac{dx}{dt}.
  • The nn-th derivative of sin and cos shifts by nπ/2n\pi/2 , a clean formula.
  • For functions involving sin⁡−1\sin^{-1} or cos⁡−1\cos^{-1}, deriving a differential equation that the function satisfies is a common JEE technique.

Next, the existence theorems for derivatives.

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