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NDA Maths · Application of Derivatives

Tangents, Rates of Change & Approximations

The derivative read geometrically (slope of the tangent), dynamically (a rate of change), and as a tool for estimating small changes via differentials.

All Application of Derivatives notesNDA Maths strategy

Why this matters

These are the most direct uses of f′(x): the slope of a tangent or normal, how fast one quantity changes with another, and a quick linear estimate of a small change. They are reliably easy marks once you read the derivative the right way.

Concept 1 of 2

Tangent and normal to a curve

Intuition

The slope of the tangent at a point is just f′(x0)f'(x_0); the normal is perpendicular, so its slope is −1/f′(x0)-1/f'(x_0). With a point and a slope, the line equations follow immediately.

Definition

At (x0,y0)(x_0,y_0) on y=f(x)y=f(x): tangent slope m=f′(x0)m=f'(x_0), tangent y−y0=m(x−x0)y-y_0=m(x-x_0); normal slope −1/m-1/m, y−y0=−1m(x−x0)y-y_0=-\tfrac1m(x-x_0). The tangent makes angle θ=tan⁡−1m\theta=\tan^{-1}m with the x-axis. A tangent is horizontal where f′=0f'=0, vertical where f′f' is undefined; parallel tangents share the same mm.

Tangent & normal at a point

mtangent=dydx∣(x1,y1)mnormal=−1dy/dxy−y1=m(x−x1)m_{\text{tangent}}=\left.\frac{dy}{dx}\right|_{(x_1,y_1)} \qquad m_{\text{normal}}=-\frac{1}{dy/dx} \qquad y-y_1=m(x-x_1)

Worked example

Find the slope of the tangent to y=x3−2xy=x^3-2x at x=1x=1.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Application of DerivativesMODERATE
The tangent to the curve x2=yx^{2}=y at (1,1)(1,1) makes an angle θ\theta with the positive direction of xx-axis. Which one of the following is correct?

[Q100 · Sep · 2021]

The normal slope is the NEGATIVE reciprocal

If the tangent slope is mm, the normal slope is −1/m-1/m — not 1/m1/m and not −m-m. Drop either the minus sign or the reciprocal and the normal line is wrong.
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Concept 2 of 2

Rates of change and small-change approximation

Intuition

A derivative is a rate: dydt\dfrac{dy}{dt} tells how fast yy changes in time, and related quantities chain together via dydt=dydxdxdt\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}. For a small input change, the derivative gives a fast linear estimate of the output change: Δy≈f′(x) Δx\Delta y\approx f'(x)\,\Delta x.

Definition

  • Related rates: differentiate the relation w.r.t. time and substitute known rates (e.g. radius growing → area's rate dAdt=2πrdrdt\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}).
  • Approximation (differentials): Δy≈dydx Δx\Delta y\approx \dfrac{dy}{dx}\,\Delta x; use it to estimate f(x+Δx)≈f(x)+f′(x)Δxf(x+\Delta x)\approx f(x)+f'(x)\Delta x.

Related rates & small-change approximation

dydt=dydx⋅dxdtΔy≈f′(x) Δxdy=f′(x) dx\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt} \qquad \Delta y\approx f'(x)\,\Delta x \qquad dy=f'(x)\,dx

Worked example

The radius of a circle grows at 33 cm/s. How fast is the area changing when r=5r=5 cm?
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Application of DerivativesEASY
The radius of a circle is increasing at the rate of 0⋅70\cdot7 cm/sec. What is the rate of increase of its circumference?

[Q72 · Apr · 2020]

Related rates need the CHAIN RULE

To get a time rate, differentiate the relation with respect to tt and chain through the variable: dAdt=dAdrdrdt\dfrac{dA}{dt}=\dfrac{dA}{dr}\dfrac{dr}{dt}. Differentiating A=πr2A=\pi r^2 as if rr were the variable gives 2πr2\pi r — the rate is 2πr drdt2\pi r\,\dfrac{dr}{dt}, not 2πr2\pi r.
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Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Tangent and normal to a curve

    Tangent & normal at a point

    mtangent=dydx∣(x1,y1)mnormal=−1dy/dxy−y1=m(x−x1)m_{\text{tangent}}=\left.\frac{dy}{dx}\right|_{(x_1,y_1)} \qquad m_{\text{normal}}=-\frac{1}{dy/dx} \qquad y-y_1=m(x-x_1)
  • Rates of change and small-change approximation

    Related rates & small-change approximation

    dydt=dydx⋅dxdtΔy≈f′(x) Δxdy=f′(x) dx\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt} \qquad \Delta y\approx f'(x)\,\Delta x \qquad dy=f'(x)\,dx

Watch out for (2)

Drill every past-year question on this subtopic

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