Q.Find differential dy for each of the following functions:
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.
Differentiate each y by the product/chain rule and multiply by dx.
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(i) Quotient/product rule on (1−2x)3(3−4x)−1.
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(ii) Chain rule on a power of (3+sin2x).
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(iii) Product + chain rule on e(⋅)cos(⋅).
(i) dy=(3−4x)22(8x−7)(1−2x)2dx (ii) dy=3(3+sin2x)1/34cos2xdx (iii) dy=ex2−5x+7[(2x−5)cos(x2−1)−2xsin(x2−1)]dx
Each differential is dy=y′(x)dx; compute y′ by the product/quotient/chain rules as appropriate.
Part (i): y=3−4x(1−2x)3. Write y=uv with u=(1−2x)3, v=(3−4x)−1.
u′=3(1−2x)2(−2)=−6(1−2x)2. v′=−(3−4x)−2(−4)=(3−4x)24.
By the product rule, y′=u′v+uv′=3−4x−6(1−2x)2+(3−4x)24(1−2x)3=(3−4x)2−6(1−2x)2(3−4x)+4(1−2x)3.
Factor (1−2x)2 from the numerator: −6(3−4x)+4(1−2x)=−18+24x+4−8x=16x−14=2(8x−7).
So y′=(3−4x)22(8x−7)(1−2x)2, hence dy=(3−4x)22(8x−7)(1−2x)2dx.
Part (ii): y=(3+sin2x)2/3. By the chain rule, y′=32(3+sin2x)−1/3⋅(2cos2x)=3(3+sin2x)1/34cos2x.
So dy=3(3+sin2x)1/34cos2xdx.
Part (iii): y=ex2−5x+7cos(x2−1). Product rule with u=ex2−5x+7,v=cos(x2−1):
u′=ex2−5x+7(2x−5). v′=−sin(x2−1)(2x).
y′=u′v+uv′=ex2−5x+7(2x−5)cos(x2−1)−2xex2−5x+7sin(x2−1)=ex2−5x+7[(2x−5)cos(x2−1)−2xsin(x2−1)].
So dy=ex2−5x+7[(2x−5)cos(x2−1)−2xsin(x2−1)]dx.
(i) dy=(3−4x)22(8x−7)(1−2x)2dx (ii) dy=3(3+sin2x)1/34cos2xdx (iii) dy=ex2−5x+7[(2x−5)cos(x2−1)−2xsin(x2−1)]dx
- Forgetting the inner-function derivative (chain rule factor) when differentiating (3+sin2x)2/3 or cos(x2−1)
- Sign slip differentiating (3−4x)−1, dropping the minus sign from the outer power rule
- 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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