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Exercise 8.1 · Q2

Q.Use the linear approximation to find approximate values of

(i) (123)2/3(123)^{2/3}
(ii) 154\sqrt[4]{15}
(iii) 263\sqrt[3]{26}
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Concept understanding — Linear Approximation and Differentials

A nonlinear function is generally hard to evaluate exactly, but near any one point its graph looks almost like a straight line -- the tangent line at that point. This is the idea behind linear approximation.

Definition (Linear Approximation). Let f:(a,b)→Rf:(a,b)\to\mathbb R be differentiable and x0∈(a,b)x_0\in(a,b). The linear approximation LL of ff at x0x_0 is

L(x)=f(x0)+f′(x0)(x−x0),∀ x∈(a,b).L(x) = f(x_0) + f'(x_0)(x-x_0), \qquad \forall\, x\in(a,b).

This is exactly the equation of the tangent line to y=f(x)y=f(x) at (x0,f(x0))(x_0,f(x_0)). Because ff is differentiable, f(x0+Δx)≈L(x0+Δx)=f(x0)+f′(x0)Δxf(x_0+\Delta x)\approx L(x_0+\Delta x) = f(x_0)+f'(x_0)\Delta x when Δx\Delta x is small, and the error f(x)−L(x)f(x)-L(x) shrinks to 00 faster than x→x0x\to x_0 (it is o(x−x0)o(x-x_0), a consequence of the definition of the derivative as a limit).

Tip

To linearly approximate an "ugly" value like 27.23\sqrt[3]{27.2} or (123)2/3(123)^{2/3}: pick a nearby point x0x_0 where the function and its derivative are easy to evaluate exactly (a perfect cube, a perfect fourth power, a multiple of 1010, ...), compute f(x0)f(x_0) and f′(x0)f'(x_0), and plug into L(x0+Δx)=f(x0)+f′(x0)ΔxL(x_0+\Delta x)=f(x_0)+f'(x_0)\Delta x.

The differential. Writing Δx=dx\Delta x = dx, the tangent-line increment is called the differential of ff:

df=f′(x) dx,equivalently df=f′(x)Δx.df = f'(x)\,dx, \qquad \text{equivalently } df = f'(x)\Delta x.

Geometrically, Δf=f(x+dx)−f(x)\Delta f = f(x+dx)-f(x) is the actual rise along the curve, while df=f′(x)dxdf=f'(x)dx is the rise along the tangent line; for small dxdx, Δf≈df\Delta f\approx df, but the two are generally not equal (only exactly equal when ff is itself linear, f(x)=mx+cf(x)=mx+c). dfdf is a function of two independent quantities, xx and dxdx -- not of xx alone the way the derivative is.

Differentials of standard functions follow directly from the derivative rules, e.g. d(xn)=nxn−1dxd(x^n)=nx^{n-1}dx, d(sin⁡x)=cos⁡x dxd(\sin x)=\cos x\,dx, d(ex)=exdxd(e^x)=e^x dx, d(log⁡x)=1xdxd(\log x)=\dfrac1x dx. The algebraic properties of differentials mirror differentiation exactly:

  • d(c)=0d(c)=0 for a constant cc;  d(x)=dx\ d(x)=dx.
  • d(cf)=c dfd(cf) = c\,df.
  • d(f±g)=df±dgd(f\pm g) = df\pm dg.
  • Product rule: d(fg)=f dg+g dfd(fg) = f\,dg + g\,df.
  • Quotient rule: d ⁣(fg)=g df−f dgg2d\!\left(\dfrac fg\right) = \dfrac{g\,df-f\,dg}{g^2}, g≠0g\ne0.
  • Chain rule: if h=f∘gh=f\circ g, dh=f′(g(x)) g′(x) dxdh = f'(g(x))\,g'(x)\,dx.
  • d(ef(x))=ef(x)f′(x) dxd(e^{f(x)}) = e^{f(x)}f'(x)\,dx;  d(log⁡f(x))=f′(x)f(x) dx\ d(\log f(x)) = \dfrac{f'(x)}{f(x)}\,dx (for f(x)>0f(x)>0).

Extending to several variables. For F:A→RF:A\to\mathbb R, A⊂R2A\subset\mathbb R^2 open, and (x0,y0)∈A(x_0,y_0)\in A, the linear approximation is

F(x,y)≈F(x0,y0)+∂F∂x∣(x0,y0)(x−x0)+∂F∂y∣(x0,y0)(y−y0),F(x,y) \approx F(x_0,y_0) + \frac{\partial F}{\partial x}\Big|_{(x_0,y_0)}(x-x_0) + \frac{\partial F}{\partial y}\Big|_{(x_0,y_0)}(y-y_0),

and the differential is dF=∂F∂xdx+∂F∂ydydF = \dfrac{\partial F}{\partial x}dx + \dfrac{\partial F}{\partial y}dy, with dx=Δx, dy=Δydx=\Delta x,\ dy=\Delta y. Geometrically this is the tangent plane to z=F(x,y)z=F(x,y) at (x0,y0)(x_0,y_0) -- exactly as the one-variable linear approximation was a tangent line. The same pattern extends to three variables:

F(x,y,z)≈F(x0,y0,z0)+Fx(x0,y0,z0)(x−x0)+Fy(x0,y0,z0)(y−y0)+Fz(x0,y0,z0)(z−z0),F(x,y,z)\approx F(x_0,y_0,z_0)+F_x(x_0,y_0,z_0)(x-x_0)+F_y(x_0,y_0,z_0)(y-y_0)+F_z(x_0,y_0,z_0)(z-z_0),

dF=Fx dx+Fy dy+Fz dz.dF = F_x\,dx + F_y\,dy + F_z\,dz.

Note

A linear approximation problem always has three ingredients: the base point (where the function is easy), the function value there, and the derivative(s) there. Get all three right and the rest is substitution.

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