Chapter Summary and Practice

Problem
Construct a function continuous everywhere but not differentiable at exactly two real inputs.

1. 1.Take f(x)=|x|+|x−1|. Each modulus function is continuous everywhere, so their sum is continuous everywhere.
2. 2.For x<0 the expression is −x+(1−x)=1−2x and its slope is −2. For 0<x<1 it is x+(1−x)=1 and its slope is 0.
3. 3.For x>1 it is x+(x−1)=2x−1 and its slope is 2.
4. 4.At zero the left and right slopes are −2 and 0; at 1 they are 0 and 2. The derivatives fail at these two inputs.
5. 5.All other inputs lie in an open interval where the function is linear. Thus zero and 1 are exactly the two nondifferentiable points, and continuity holds even there.

Problem
For f(x)=|x|³, find f″(x) for all real x, including zero.

1. 1.For x>0, f=x³, so f′=3x² and f″=6x. For x<0, f=−x³, so f′=−3x² and f″=−6x.
2. 2.At zero the first difference quotient is |h|³/h. Its magnitude is h², which tends to zero, so f′(0)=0.
3. 3.For h≠0, f′(h)=3h|h|. The difference quotient for f′ at zero is [f′(h)−f′(0)]/h=3|h|, which also tends to zero.
4. 4.Therefore f″(0)=0 and f″(x)=6|x| for every real x. The modulus itself has a corner, but cubing its magnitude changes the behaviour enough to allow two derivatives.

Problem
Differentiate y=√(3x+2)+1/√(2x²+4).

1. 1.The second square-root denominator is always positive. The first root requires 3x+2≥0, so the function domain is x≥−2/3.
2. 2.Write y=(3x+2)⁽¹⁄²⁾+(2x²+4)⁽⁻¹⁄²⁾. The power form makes the chain-rule multipliers explicit.
3. 3.Differentiating gives y′=(1/2)(3x+2)⁽⁻¹⁄²⁾·3−(1/2)(2x²+4)⁽⁻³⁄²⁾·4x.
4. 4.Hence y′=3/[2√(3x+2)]−2x/(2x²+4)⁽³⁄²⁾, valid for x>−2/3. The left endpoint belongs to the function domain but not to this finite derivative domain.

Problem
Differentiate y=log₇(log x).

1. 1.The inner logarithm requires x>0, and the outer logarithm requires log x>0. Together these give x>1.
2. 2.Change base: y=log(log x)/log7. The denominator log7 is a nonzero constant.
3. 3.Differentiate the numerator using the logarithmic chain rule: d[log(log x)]/dx=1/[x log x].
4. 4.Thus y′=1/[x log7 log x] for x>1. Neither a base-7 logarithm nor a repeated logarithm requires a new differentiation principle.

Problem
Differentiate f(x)=cos⁻¹(sin x) on its differentiable intervals.

1. 1.The function is defined for every real x because sin x lies in [−1,1]. At points where cos x≠0, the chain rule gives f′=−cos x/√(1−sin²x).
2. 2.Since √(cos²x)=|cos x|, the derivative is −cos x/|cos x|.
3. 3.On −π/2<x<π/2, cosine is positive, so f′=−1 and f(x)=π/2−x. On π/2<x<3π/2, cosine is negative, so f′=1 and f(x)=x−π/2.
4. 4.The pattern repeats with period 2π. At x=π/2+kπ the one-sided slopes disagree, giving corners and no two-sided derivative.
5. 5.Thus −1 is a correct derivative on particular intervals, not a global answer for this function.

Problem
Differentiate f(x)=tan⁻¹[sin x/(1+cos x)].

1. 1.The denominator excludes x=(2k+1)π. On the remaining intervals, sin x/(1+cos x)=tan(x/2).
2. 2.For −π<x<π, x/2 lies in (−π/2,π/2), so f(x)=x/2 and f′=1/2.
3. 3.More generally on (2k−1)π<x<(2k+1)π, the principal value is f(x)=x/2−kπ. The constant shift changes the value but not the derivative.
4. 4.Therefore f′=1/2 at every allowed input. The values cannot be identified globally with x/2, and the excluded inputs remain excluded.

Problem
Differentiate f(x)=sin⁻¹[2ˣ⁺¹/(1+4ˣ)].

1. 1.Put z=2ˣ>0. The argument is 2z/(1+z²), which is at most 1 because (z−1)²≥0, and is positive. Thus f is defined for every real x.
2. 2.Let θ=tan⁻¹z, so 0<θ<π/2 and the argument is sin2θ. If x≤0 then z≤1 and 2θ≤π/2, giving f=2tan⁻¹(2ˣ).
3. 3.If x≥0, 2θ≥π/2. Inverse sine returns the reflected principal angle, giving f=π−2tan⁻¹(2ˣ). The two expressions agree at x=0 with value π/2.
4. 4.For x<0, f′=2ˣ⁺¹log2/(1+4ˣ). For x>0, f′=−2ˣ⁺¹log2/(1+4ˣ).
5. 5.At zero the left derivative is log2 and the right derivative is −log2. The function is continuous there but has no two-sided derivative. This connects principal values directly to the continuity–differentiability distinction.

Problem
Find f′(x) for f(x)=(sin x)⁽ˢⁱⁿ ˣ⁾, 0<x<π.

1. 1.On this interval sin x>0, so the function is positive and log f=sin x·log(sin x).
2. 2.Differentiate: f′/f=cos x log(sin x)+sin x·[cos x/sin x].
3. 3.Simplify the second term to cos x, giving f′/f=(1+log(sin x))cos x.
4. 4.Thus f′=(sin x)⁽ˢⁱⁿ ˣ⁾(1+log(sin x))cos x. The stated interval justifies all logarithms; it is not valid to claim a real variable-power definition for every positive x without checking the sine base.

Problem
For a>0, let y=a⁽ᵗ⁺¹⁄ᵗ⁾ and x=(t+1/t)ᵃ. Find dy/dx using t>0 and t≠1.

1. 1.The restriction t>0 ensures that t+1/t is positive, so its arbitrary real power a is defined. Write u=t+1/t, with du/dt=1−1/t².
2. 2.The coordinate derivatives are dy/dt=aᵘlog a·(1−1/t²) and dx/dt=a uᵃ⁻¹(1−1/t²).
3. 3.For t>0,t≠1, the horizontal rate is nonzero. Dividing gives dy/dx=aᵘlog a/[a uᵃ⁻¹], with u=t+1/t.
4. 4.At t=1 both rates vanish, so this division does not itself establish the derivative. The relation may instead be examined through the positive variable u. A negative t would make u negative, which is not a valid base for every arbitrary positive real a.

Problem
Differentiate sin²x with respect to exp(cos x), using a point where sin x≠0.

1. 1.Set u=sin²x and v=exp(cos x). Their derivatives with respect to x are du/dx=2sin x cos x and dv/dx=−sin x·exp(cos x).
2. 2.Where sin x≠0 the denominator rate is nonzero, so du/dv=(2sin x cos x)/(−sin x·exp(cos x)).
3. 3.Cancel the nonzero sine factor to obtain du/dv=−2cos x/exp(cos x).
4. 4.The cancellation records a formula on the inputs where the rate ratio was justified. If an exceptional input is required, examine the relationship between u and v separately rather than divide 0 by 0.

Problem
For x=a(cos t+t sin t), y=a(sin t−t cos t), a>0, find y′ and y″ where t≠0 and cos t≠0.

1. 1.Differentiate the coordinates: dx/dt=a[−sin t+sin t+t cos t]=at cos t, and dy/dt=a[cos t−cos t+t sin t]=at sin t.
2. 2.The first slope is y′=tan t at the stated parameters.
3. 3.Differentiate this slope with respect to t to get sec²t.
4. 4.Convert to differentiation with respect to x: y″=sec²t/(at cos t)=1/(at cos³t). The second division is essential.

Problem
For (x−a)²+(y−b)²=c², c>0, examine [1+(y′)²]⁽³⁄²⁾/y″ on a smooth branch with y≠b.

1. 1.Implicit differentiation gives (x−a)+(y−b)y′=0, hence y′=−(x−a)/(y−b).
2. 2.Using the circle equation, 1+(y′)²=[(y−b)²+(x−a)²]/(y−b)²=c²/(y−b)².
3. 3.Differentiate (x−a)+(y−b)y′=0 again: 1+(y′)²+(y−b)y″=0. Thus y″=−c²/(y−b)³.
4. 4.The numerator [1+(y′)²]⁽³⁄²⁾ is c³/|y−b|³. Dividing gives −c·sign(y−b): it is −c on an upper branch and +c on a lower branch.
5. 5.The result is constant on each such branch and independent of a,b. If the denominator were |y″| instead, the ratio would be the positive constant c. The absolute value and the branch distinction explain the sign.

Problem
For y=exp(a cos⁻¹x), prove (1−x²)y″−xy′−a²y=0 on −1<x<1.

1. 1.The chain rule gives y′=−ay/√(1−x²). Rewrite this as √(1−x²)y′=−ay.
2. 2.Differentiate the rewritten identity: √(1−x²)y″−x y′/√(1−x²)=−ay′.
3. 3.Multiply by √(1−x²): (1−x²)y″−xy′=−a√(1−x²)y′.
4. 4.Use the first identity again on the right: −a√(1−x²)y′=a²y. Moving this term to the left proves the result.
5. 5.Although the original function is defined at x=±1, the derivative-based proof and identity are asserted on the open interval where the inverse-cosine derivative is finite.