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Methods of Differentiation — chapter notes

6 sections 4 min readBeta

1. Implicit functions

When xx and yy are mixed, differentiate every term with respect to xx, treating yy as a function of xx:

  • ddx(yn)=nyn−1dydx\dfrac{d}{dx}(y^n) = ny^{n - 1}\dfrac{dy}{dx}, ddx(xy)=y+xdydx\dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx}, ddxey=eydydx\dfrac{d}{dx}e^{y} = e^{y}\dfrac{dy}{dx}.
  • Collect the dydx\dfrac{dy}{dx} terms and divide. For F(x,y)=0F(x, y) = 0 the result is −FxFy-\dfrac{F_x}{F_y}, where FxF_x treats yy as a constant.
  • Writing dydx=ND\dfrac{dy}{dx} = \dfrac{N}{D}: a horizontal tangent needs N=0N = 0, a vertical tangent D=0D = 0 (and the point must lie on the curve).
Four steps, every time.
  • upper branch of x2+2xy+3y2=6x^2 + 2xy + 3y^2 = 6
  • tangent at (1,1)(1, 1)
Worked example 1: slope $-\dfrac12$ at $(1, 1)$.
Read the fraction, not just its value.
TangentCondition on dydx=ND\dfrac{dy}{dx} = \dfrac{N}{D}Example on the circle
HorizontalN=0N = 0, D≠0D \ne 0(−2,6)(-2, 6) and (−2,−4)(-2, -4)
VerticalD=0D = 0, N≠0N \ne 0(3,1)(3, 1) and (−7,1)(-7, 1)

Formulas

  • Implicit slope
    dydx=−FxFy\dfrac{dy}{dx} = -\dfrac{F_x}{F_y}

Key terms

Implicit function

A function defined by an equation in which y is not isolated.

Horizontal tangent

A tangent where the slope is zero.

2. Parametric form

If x=f(t)x = f(t) and y=g(t)y = g(t), then dydx=g′(t)f′(t)\dfrac{dy}{dx} = \dfrac{g'(t)}{f'(t)} wherever f′(t)≠0f'(t) \ne 0. The answer may stay in terms of tt.

Useful pairs: x=acos⁡tx = a\cos t, y=bsin⁡ty = b\sin t gives −bacot⁡t-\dfrac{b}{a}\cot t; x=at2x = at^2, y=2aty = 2at gives 1t\dfrac{1}{t}. A parameter can describe curves that cross themselves, which no single formula y=f(x)y = f(x) can.

  • x=t2−1, y=t3−tx = t^2 - 1,\ y = t^3 - t
A curve that is easier to describe with a parameter.
Two values of $t$ give the same point $(0, 0)$: the curve crosses itself.
tt−1.5-1.5−1-1−0.5-0.5000.50.5111.51.5
x=t2−1x = t^2 - 11.251.2500−0.75-0.75−1-1−0.75-0.75001.251.25
y=t3−ty = t^3 - t−1.875-1.875000.3750.37500−0.375-0.375001.8751.875
Computed by dividing the two rates.
StepResult
dxdt\dfrac{dx}{dt}4t4t
dydt\dfrac{dy}{dt}44
dydx\dfrac{dy}{dx}1t\dfrac{1}{t}
At t=2t = 212\dfrac{1}{2} at the point (8,8)(8, 8)

Formulas

  • Parametric slope
    dydx=dy/dtdx/dt\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}

Key terms

Parameter

A third variable, such as t or θ, in terms of which both x and y are given.

3. One function with respect to another

To differentiate u(x)u(x) with respect to v(x)v(x), divide: dudv=du/dxdv/dx\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}, where dvdx≠0\dfrac{dv}{dx} \ne 0.

Often a substitution shows the answer directly: with x=sin⁡θx = \sin\theta, cos⁡−1(1−2x2)=2θ=2sin⁡−1x\cos^{-1}(1 - 2x^2) = 2\theta = 2\sin^{-1} x for 0<x<10 < x < 1, so its derivative with respect to sin⁡−1x\sin^{-1} x is 22. Always check the range of the angle before cancelling an inverse function.

Both functions depend on the same $x$.
A numerical check of the answer $-2\cos x$.
xxπ6\dfrac{\pi}{6}π4\dfrac{\pi}{4}π3\dfrac{\pi}{3}
ratio of derivatives (numeric)−1.7321-1.7321−1.4142-1.4142−1-1
−2cos⁡x-2\cos x−1.7321-1.7321−1.4142-1.4142−1-1

Formulas

  • Relative derivative
    dudv=du/dxdv/dx\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}

Key terms

Derivative of u with respect to v

The rate of change of u per unit change of v, found as (du/dx)/(dv/dx).

4. Logarithmic differentiation

For y=f(x)g(x)y = f(x)^{g(x)} (with f(x)>0f(x) > 0): log⁡y=g(x)log⁡f(x)\log y = g(x)\log f(x), so dydx=y[g′(x)log⁡f(x)+g(x)f′(x)f(x)]\dfrac{dy}{dx} = y\left[g'(x)\log f(x) + g(x)\dfrac{f'(x)}{f(x)}\right].

  • ddxxx=xx(1+log⁡x)\dfrac{d}{dx}x^{x} = x^{x}(1 + \log x).
  • Sums: split y=u+vy = u + v and differentiate each term by its own method.
  • Long products and quotients: take logs to turn them into sums, e.g. 1ydydx=∑nix−ai\dfrac{1}{y}\dfrac{dy}{dx} = \sum \dfrac{n_i}{x - a_i} for y=∏(x−ai)niy = \prod (x - a_i)^{n_i}.
  • Towers: xxxx^{x^{x}} needs logs twice; the endless tower satisfies y=xyy = x^{y}, giving dydx=y2x(1−ylog⁡x)\dfrac{dy}{dx} = \dfrac{y^2}{x(1 - y\log x)}.
The last step is the one students forget.
Logs turn powers into products and products into sums.
FunctionWhy logs help
xxx^{x}, (log⁡x)x(\log x)^{x}variable base and variable exponent
(x+1)5(x−2)3(x+4)2\dfrac{(x + 1)^5(x - 2)^3}{(x + 4)^2}a long product and quotient becomes a sum
xxxx^{x^{x}}a tower: take logs twice
Split first, then take logs term by term.
TermTreat it asDerivative
u=xxu = x^{x}log⁡u=xlog⁡x\log u = x\log xxx(1+log⁡x)x^{x}(1 + \log x)
v=2xv = 2^{x}exponential rule2xlog⁡22^{x}\log 2
y=u+vy = u + vadd the twoxx(1+log⁡x)+2xlog⁡2x^{x}(1 + \log x) + 2^{x}\log 2
The endless tower settles at $y \approx 1.258$, and then $y = 1.2^{y}$.
Step kkyk=1.2 yk−1y_k = 1.2^{\,y_{k - 1}}
111.21.2
221.24461.2446
331.25471.2547
441.2571.257
551.25761.2576
661.25771.2577

Formulas

  • Variable power
    ddx(xx)=xx(1+log⁡x)\dfrac{d}{dx}\left(x^{x}\right) = x^{x}(1 + \log x)

Key terms

Logarithmic differentiation

Taking logarithms of both sides before differentiating, then multiplying back by y.

Variable power

A power whose base and exponent both depend on x.

5. Second-order derivatives

d2ydx2=ddx(dydx)\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right), also written y′′y'' or y2y_2. Its sign tells how the slope changes: positive, the curve bends upwards; negative, it bends downwards. (Derivatives beyond the second are not in the ISC 2027 syllabus.)

  • Parametric: d2ydx2=ddt(dydx)÷dxdt\dfrac{d^2y}{dx^2} = \dfrac{d}{dt}\left(\dfrac{dy}{dx}\right) \div \dfrac{dx}{dt}.
  • Implicit: differentiate the equation for y′y' again and substitute y′y'; e.g. for x2+y2=r2x^2 + y^2 = r^2, y′′=−r2y3y'' = -\dfrac{r^2}{y^3}.
  • Relations: to prove a relation such as y′′−4y′+5y=0y'' - 4y' + 5y = 0, find y′y' and y′′y'', then substitute; or clear denominators first and differentiate the product.
  • y=x3−3xy = x^3 - 3x
$\dfrac{d^2y}{dx^2} = 6x$: negative on the left (bends down), positive on the right (bends up).
The extra division by $\dfrac{dx}{dt}$ is the step most often missed.
Product rule twice, then collect.
StepResult
y′y'ex(cos⁡x−sin⁡x)e^{x}(\cos x - \sin x)
y′′y''ex(cos⁡x−sin⁡x)+ex(−sin⁡x−cos⁡x)e^{x}(\cos x - \sin x) + e^{x}(-\sin x - \cos x)
Simplified−2exsin⁡x-2e^{x}\sin x

Formulas

  • Parametric second derivative
    d2ydx2=ddt(dydx)÷dxdt\dfrac{d^2y}{dx^2} = \dfrac{d}{dt}\left(\dfrac{dy}{dx}\right) \div \dfrac{dx}{dt}

Key terms

Second-order derivative

The derivative of the first derivative.

6. Choosing the method

Read the form first: mixed xx and yy means implicit; both given by a parameter means parametric; "with respect to" another function means divide two derivatives; a variable base with a variable exponent, or a long product, means logarithms. Several methods often combine in one question, for example the endless tower is both implicit and logarithmic.

Read the form first, then choose.
You seeUse
xx and yy mixed, yy not isolatedimplicit differentiation
xx and yy both given by tt or θ\thetaparametric: dy/dtdx/dt\dfrac{dy/dt}{dx/dt}
"differentiate uu with respect to vv"du/dxdv/dx\dfrac{du/dx}{dv/dx}
variable in base and exponentlogarithmic differentiation (Part 2)

Key terms

Method choice

Deciding, from the form of the function, which differentiation method applies.

Exam tips

How ISC marks are awarded

  • Implicit slope (2–3 marks): differentiation with the product and chain rules, collecting terms, then the value at the point.
  • Parametric (2–3 marks): both rates, the quotient, and the extra division for a second derivative.
  • Logarithmic (2–3 marks): the log line, differentiation, multiplying back by yy.
  • Proving a second-order relation (3–5 marks): y′y', y′′y'', substitution and simplification on separate lines.
  • Section D (5 marks): parts such as 2 + 3; label each part.

Common mistakes

  • Forgetting dydx\dfrac{dy}{dx} on a yy-term, or the product rule on xyxy.
  • Inverting the parametric quotient.
  • Taking the log of a sum.
  • Forgetting to multiply back by yy.
  • Treating d2ydx2\dfrac{d^2y}{dx^2} as ddt(dydx)\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right) without dividing by dxdt\dfrac{dx}{dt}.