Q.Find df for f(x)=x2+3x and evaluate it for
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.
df=f′(x)dx=(2x+3)dx for f(x)=x2+3x.
-
(i) x=2,dx=0.1: (7)(0.1).
-
(ii) x=3,dx=0.02: (9)(0.02).
(i) df=0.7 (ii) df=0.18
f(x)=x2+3x⇒f′(x)=2x+3, so df=(2x+3)dx; substitute each given (x,dx) pair.
Step 1. General differential. f′(x)=2x+3, so df=(2x+3)dx.
Step 2. Part (i): x=2, dx=0.1. df=(2(2)+3)(0.1)=(4+3)(0.1)=7(0.1)=0.7.
Step 3. Part (ii): x=3, dx=0.02. df=(2(3)+3)(0.02)=(6+3)(0.02)=9(0.02)=0.18.
(i) df=0.7 (ii) df=0.18
- Using x alone in f′(x) without multiplying by the given dx
- Substituting the wrong x-value from the other sub-part
- 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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