Q.Find a linear approximation for the following functions at the indicated points.
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)→R be differentiable and x0∈(a,b). The linear approximation L of f at x0 is
L(x)=f(x0)+f′(x0)(x−x0),∀x∈(a,b).
This is exactly the equation of the tangent line to y=f(x) at (x0,f(x0)). Because f is differentiable, f(x0+Δx)≈L(x0+Δx)=f(x0)+f′(x0)Δx when Δx is small, and the error f(x)−L(x) shrinks to 0 faster than x→x0 (it is o(x−x0), a consequence of the definition of the derivative as a limit).
To linearly approximate an "ugly" value like 327.2 or (123)2/3: pick a nearby point x0 where the function and its derivative are easy to evaluate exactly (a perfect cube, a perfect fourth power, a multiple of 10, ...), compute f(x0) and f′(x0), and plug into L(x0+Δx)=f(x0)+f′(x0)Δx.
The differential. Writing Δx=dx, the tangent-line increment is called the differential of f:
df=f′(x)dx,equivalently df=f′(x)Δx.
Geometrically, Δf=f(x+dx)−f(x) is the actual rise along the curve, while df=f′(x)dx is the rise along the tangent line; for small dx, Δf≈df, but the two are generally not equal (only exactly equal when f is itself linear, f(x)=mx+c). df is a function of two independent quantities, x and dx -- not of x alone the way the derivative is.
Differentials of standard functions follow directly from the derivative rules, e.g. d(xn)=nxn−1dx, d(sinx)=cosxdx, d(ex)=exdx, d(logx)=x1dx. The algebraic properties of differentials mirror differentiation exactly:
- d(c)=0 for a constant c; d(x)=dx.
- d(cf)=cdf.
- d(f±g)=df±dg.
- Product rule: d(fg)=fdg+gdf.
- Quotient rule: d(gf)=g2gdf−fdg, g=0.
- Chain rule: if h=f∘g, dh=f′(g(x))g′(x)dx.
- d(ef(x))=ef(x)f′(x)dx; d(logf(x))=f(x)f′(x)dx (for f(x)>0).
Extending to several variables. For F:A→R, A⊂R2 open, and (x0,y0)∈A, the linear approximation is
F(x,y)≈F(x0,y0)+∂x∂F(x0,y0)(x−x0)+∂y∂F(x0,y0)(y−y0),
and the differential is dF=∂x∂Fdx+∂y∂Fdy, with dx=Δx, dy=Δy. Geometrically this is the tangent plane to z=F(x,y) at (x0,y0) -- 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),
dF=Fxdx+Fydy+Fzdz.
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.
Compute f(x0),f′(x0) and substitute into L(x)=f(x0)+f′(x0)(x−x0).
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(i) f′(x)=3x2−5: f(2)=10,f′(2)=7⇒L=7x−4.
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(ii) g′(x)=x/x2+9: g(−4)=5,g′(−4)=−54⇒L=−54x+59.
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(iii) h′(x)=1/(x+1)2: h(1)=21,h′(1)=41⇒L=4x+1.
(i) L(x)=7x−4 (ii) L(x)=−54x+59 (iii) L(x)=41x+41
Each part needs f(x0) and f′(x0) at the given point, then direct substitution into L(x)=f(x0)+f′(x0)(x−x0).
Part (i): f(x)=x3−5x+12, x0=2.
f(2)=8−10+12=10. f′(x)=3x2−5, so f′(2)=12−5=7.
L(x)=10+7(x−2)=10+7x−14=7x−4.
Part (ii): g(x)=x2+9, x0=−4.
g(−4)=16+9=25=5. g′(x)=2x2+92x=x2+9x, so g′(−4)=5−4.
L(x)=5−54(x+4)=5−54x−516=525−16−54x=59−54x.
Part (iii): h(x)=x+1x, x0=1.
h(1)=21. By the quotient rule, h′(x)=(x+1)2(x+1)(1)−x(1)=(x+1)21, so h′(1)=41.
L(x)=21+41(x−1)=41x+21−41=41x+41=4x+1.
(i) L(x)=7x−4 (ii) L(x)=−54x+59 (iii) L(x)=4x+1
Linear approximation — evaluate f(x0) and f′(x0), substitute into L(x)=f(x0)+f′(x0)(x−x0)
- Forgetting the chain rule's factor of 2x inside the square root in part (ii) before it cancels with the 21 power
- Not fully expanding L(x) into slope-intercept form, leaving it as an unsimplified point-slope expression
- CBSE 2025Set ANNUAL1 markMCQQ.If f(x)>0 for all x and g(x)=log(f(x)), then dg is :(a) f(x)1dx(b) f(x)1f′(x)dx(c) x1dx(d) x1f(x)dx
›Reveal solutionSolution
Differentiating the composite function log(f(x)) by the chain rule gives f′(x)/f(x), and the differential is this derivative times dx.
- g(x)=log(f(x)), with f(x)>0 so the logarithm is defined.
- By the chain rule, g′(x)=dxdlog(f(x))=f(x)1⋅f′(x) (derivative of logu is u1⋅dxdu with u=f(x)).
- The differential of g is defined as dg=g′(x)dx.
- Substituting: dg=f(x)1f′(x)dx.
✓Final answer(b) f(x)1f′(x)dx
- CBSE 2024Set ANNUAL1 markMCQQ.If f(x)=x+1x, then its differential is given by :(a) x+11dx(b) (x+1)2−1dx(c) x+1−1dx(d) (x+1)21dx
›Reveal solutionSolution
Applying the quotient rule to x/(x+1) gives a clean derivative, from which the differential follows immediately.
- f(x)=x+1x. By the quotient rule, f′(x)=(x+1)2(1)(x+1)−x(1)=(x+1)2x+1−x=(x+1)21.
- The differential is df=f′(x)dx=(x+1)21dx.
✓Final answer(d) (x+1)21dx
- CBSE 2022Set ANNUAL1 markMCQQ.If f(x)=x+1x, then its differential is :(a) x+11dx(b) (x+1)2−1dx(c) x+1−1dx(d) (x+1)21dx
›Reveal solutionSolution
Differentiating f(x)=x+1x by the quotient rule gives f′(x)=(x+1)21, so dy=(x+1)21dx.
- Let y=f(x)=x+1x.
- By the quotient rule, dxdy=(x+1)2(x+1)⋅dxd(x)−x⋅dxd(x+1).
- This gives dxdy=(x+1)2(x+1)(1)−x(1)=(x+1)2x+1−x=(x+1)21.
- The differential of y is defined as dy=f′(x)dx.
- Substituting, dy=(x+1)21dx.
✓Final answerdy=(x+1)21dx — option (d).
- CBSE 2018Set ANNUAL1 markMCQQ.The differential of y if y=x4+x2+1 is :(a) 21(4x3+2x)−21(b) 21(4x3+2x)−21dx(c) 21(x4+x2+1)−21(4x3+2x)(d) 21(x4+x2+1)−21(4x3+2x)dx
›Reveal solutionSolution
Applying the chain rule to y=x4+x2+1 and multiplying by dx gives the differential dy=21(x4+x2+1)−1/2(4x3+2x)dx.
- Write y=(x4+x2+1)1/2.
- By the chain rule, dxdy=21(x4+x2+1)−21⋅dxd(x4+x2+1).
- Compute the inner derivative: dxd(x4+x2+1)=4x3+2x.
- So dxdy=21(x4+x2+1)−21(4x3+2x).
- The differential of y is dy=dxdydx, i.e. dy=21(x4+x2+1)−21(4x3+2x)dx — this must include both the factor (x4+x2+1)−1/2 (not just the inner-function derivative alone) and the dx.
✓Final answerdy=21(x4+x2+1)−21(4x3+2x)dx — option (d).
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