Exercise 5.5 answers: Continuity and Differentiability

Open NCERT p. 134Matches NCERT’s answer

1. This is a product of three functions of x, so take logs first.
2. Let y = cos x . cos 2x . cos 3x. Take log of both sides: log y = log(cos x) + log(cos 2x) + log(cos 3x).
3. Differentiate both sides w.r.t. x. On the left you get (1/y)(dy/dx), by the chain rule.
4. On the right, each term gives −tan of its angle, times the derivative of that angle: −tan x, −2 tan 2x, −3 tan 3x.
5. So (1/y)(dy/dx) = −(tan x + 2 tan 2x + 3 tan 3x).
6. Multiply both sides by y to get dy/dx.

Answerdy/dx = −cos x cos 2x cos 3x (tan x + 2 tan 2x + 3 tan 3x)

Watch this explained “The second use”, 12:53 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. Let y be the given square root. Take log of both sides.
2. The square root becomes a factor of 1/2 in front: log y = (1/2)[log(x−1) + log(x−2) − log(x−3) − log(x−4) − log(x−5)].
3. Differentiate both sides. The left side gives (1/y)(dy/dx).
4. Each log(x−a) term gives 1/(x−a) on the right.
5. (1/y)(dy/dx) = (1/2)[1/(x−1) + 1/(x−2) − 1/(x−3) − 1/(x−4) − 1/(x−5)].
6. Multiply both sides by y.

Answerdy/dx = (y/2)[1/(x−1) + 1/(x−2) − 1/(x−3) − 1/(x−4) − 1/(x−5)], where y is the given square root

Watch this explained “One worked in full”, 13:58 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. Both the base (log x) and the power (cos x) depend on x, so no earlier rule applies directly. Take logs first.
2. Let y = (log x)^(cos x). Then log y = cos x · log(log x).
3. Differentiate both sides. The left gives (1/y)(dy/dx).
4. The right needs the product rule: cos x times derivative of log(log x), plus log(log x) times derivative of cos x.
5. Derivative of log(log x) is 1/(x log x); derivative of cos x is −sin x.
6. (1/y)(dy/dx) = cos x/(x log x) − sin x · log(log x).
7. Multiply both sides by y.

Answerdy/dx = (log x)^(cos x) [cos x/(x log x) − sin x · log(log x)]

Watch this explained “The input, to the sine of the input”, 16:22 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. This is a difference of two terms, so differentiate each one separately (you cannot take log of a difference).
2. First term: y₁ = xˣ. Take log: log y₁ = x log x. Differentiate: (1/y₁)(dy₁/dx) = log x + 1. So dy₁/dx = xˣ(1 + log x).
3. Second term: y₂ = 2^(sin x), a constant base to a variable power. dy₂/dx = 2^(sin x) · log 2 · cos x.
4. Subtract: dy/dx = dy₁/dx − dy₂/dx.

Answerdy/dx = xˣ(1 + log x) − 2^(sin x) · cos x · log 2

Watch this explained “What a logarithm will not do”, 19:01 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. This is a product of three power terms, so take logs first.
2. Let y = (x+3)²(x+4)³(x+5)⁴. Then log y = 2 log(x+3) + 3 log(x+4) + 4 log(x+5).
3. Differentiate both sides. The left gives (1/y)(dy/dx).
4. Each term on the right gives its power over its own bracket: 2/(x+3), 3/(x+4), 4/(x+5).
5. (1/y)(dy/dx) = 2/(x+3) + 3/(x+4) + 4/(x+5).
6. Multiply both sides by y.

Answerdy/dx = (x+3)²(x+4)³(x+5)⁴ [2/(x+3) + 3/(x+4) + 4/(x+5)]

Watch this explained “The second use”, 12:53 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. This is a sum of two terms, each a variable base to a variable power. Differentiate each term separately.
2. First term: u = (x + 1/x)ˣ. Take log: log u = x · log(x + 1/x).
3. Differentiate by the product rule: (1/u)(du/dx) = log(x + 1/x) + x · (1 − 1/x²)/(x + 1/x), using the derivative of x + 1/x, which is 1 − 1/x².
4. So du/dx = u [log(x + 1/x) + x(1 − 1/x²)/(x + 1/x)].
5. Second term: v = x^(1 + 1/x). Take log: log v = (1 + 1/x) · log x.
6. Differentiate by the product rule: (1/v)(dv/dx) = (1 + 1/x)/x − (log x)/x², using the derivative of 1 + 1/x, which is −1/x².
7. So dv/dx = v [(1 + 1/x)/x − (log x)/x²].
8. Add du/dx and dv/dx.

Answerdy/dx = (x+1/x)ˣ[log(x+1/x) + x(1−1/x²)/(x+1/x)] + x^(1+1/x)[(1+1/x)/x − (log x)/x²]

Watch this explained “What a logarithm will not do”, 19:01 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. Differentiate: (log x)ˣ + x^(log x).
2. Sum of two variable-base, variable-power terms; differentiate each one separately.
3. First term: u = (log x)ˣ. Take log: log u = x · log(log x).
4. Differentiate by the product rule: (1/u)(du/dx) = log(log x) + x · 1/(x log x) = log(log x) + 1/log x.
5. So du/dx = u [log(log x) + 1/log x].
6. Second term: v = x^(log x). Take log: log v = (log x)(log x) = (log x)².
7. Differentiate: (1/v)(dv/dx) = 2 log x · (1/x).
8. So dv/dx = v · 2 log x / x.
9. Add du/dx and dv/dx.

Answerdy/dx = (log x)ˣ[log(log x) + 1/log x] + x^(log x) · 2 log x / x

Watch this explained “What a logarithm will not do”, 19:01 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. Differentiate: (sin x)ˣ + sin⁻¹ √x.
2. The two terms need different methods, so differentiate them separately.
3. First term: u = (sin x)ˣ. Take log: log u = x · log(sin x).
4. Differentiate by the product rule: (1/u)(du/dx) = log(sin x) + x · cos x/sin x = log(sin x) + x cot x.
5. So du/dx = u [log(sin x) + x cot x].
6. Second term: v = sin⁻¹(√x). This is an ordinary chain rule, not logarithmic differentiation.
7. Derivative of sin⁻¹(w) is 1/√(1−w²), with w = √x, so 1−w² = 1−x. Derivative of √x is 1/(2√x).
8. dv/dx = 1/√(1−x) × 1/(2√x) = 1/(2√(x−x²)).
9. Add du/dx and dv/dx.

Answerdy/dx = (sin x)ˣ[log(sin x) + x cot x] + 1/(2√(x−x²))

Watch this explained “What a logarithm will not do”, 19:01 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. Differentiate: x^(sin x) + (sin x)^(cos x).
2. Sum of two variable-base, variable-power terms; differentiate each one separately.
3. First term: u = x^(sin x). Take log: log u = sin x · log x.
4. Differentiate by the product rule: (1/u)(du/dx) = cos x · log x + sin x/x.
5. So du/dx = u [cos x · log x + sin x/x].
6. Second term: v = (sin x)^(cos x). Take log: log v = cos x · log(sin x).
7. Differentiate by the product rule: (1/v)(dv/dx) = −sin x · log(sin x) + cos x · (cos x/sin x).
8. So dv/dx = v [cos²x/sin x − sin x · log(sin x)].
9. Add du/dx and dv/dx.

Answerdy/dx = x^(sin x)[cos x · log x + sin x/x] + (sin x)^(cos x)[cos²x/sin x − sin x · log(sin x)]

Watch this explained “What a logarithm will not do”, 19:01 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. The two terms need different methods, so differentiate them separately.
2. First term: u = x^(x cos x). Take log: log u = x cos x · log x.
3. Differentiate the right side, which is a product of three moving pieces (x, cos x, log x). Using the product rule twice: cos x · log x + x·(−sin x)·log x + x cos x·(1/x).
4. That simplifies to cos x(1 + log x) − x sin x log x.
5. So du/dx = u [cos x(1 + log x) − x sin x log x].
6. Second term: an ordinary quotient rule on (x²+1)/(x²−1).
7. d/dx = [2x(x²−1) − (x²+1)(2x)] / (x²−1)² = −4x/(x²−1)².
8. Add the two derivatives.

Answerdy/dx = x^(x cos x)[cos x(1 + log x) − x sin x log x] − 4x/(x² − 1)²

Watch this explained “What a logarithm will not do”, 19:01 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. Sum of two variable-base, variable-power terms; differentiate each one separately.
2. First term: u = (x cos x)ˣ. Take log: log u = x · log(x cos x) = x[log x + log(cos x)].
3. Differentiate by the product rule: (1/u)(du/dx) = log x + log(cos x) + x[1/x − tan x] = log(x cos x) + 1 − x tan x.
4. So du/dx = u [log(x cos x) + 1 − x tan x].
5. Second term: v = (x sin x)^(1/x). Take log: log v = (1/x)[log x + log(sin x)].
6. Differentiate by the quotient/product rule: (1/v)(dv/dx) = [1 − log(x sin x)]/x² + cot x/x.
7. So dv/dx = v {[1 − log(x sin x)]/x² + cot x/x}.
8. Add du/dx and dv/dx.

Answerdy/dx = (x cos x)ˣ[log(x cos x) + 1 − x tan x] + (x sin x)^(1/x){[1 − log(x sin x)]/x² + cot x/x}

Watch this explained “What a logarithm will not do”, 19:01 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. The left side is a sum, and there is no log rule for a sum, so name the two terms first: u = x^y and v = y^x. Then u + v = 1, so du/dx + dv/dx = 0.
2. For u = x^y: log u = y·log x. Differentiate with respect to x (product rule on the right): (1/u)·du/dx = (dy/dx)·log x + y/x. So du/dx = x^y·log x·(dy/dx) + y·x^y−1.
3. For v = y^x: log v = x·log y. Differentiate: (1/v)·dv/dx = log y + (x/y)·(dy/dx). So dv/dx = y^x·log y + x·y^x−1·(dy/dx).
4. Add the two and set the sum equal to 0: [x^y·log x + x·y^x−1]·(dy/dx) + y·x^y−1 + y^x·log y = 0.
5. Solve for dy/dx by moving the terms without dy/dx to the right and dividing by the bracket.

Answerdy/dx = −[y·x^y−1 + y^x·log y] ÷ [x^y·log x + x·y^x−1]

Watch this explained “What a logarithm will not do”, 19:01 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. Find dy/dx when yˣ = xʸ.
2. Take the natural log of both sides: x·log y = y·log x.
3. Differentiate both sides with respect to x, using the product rule on each side.
4. Left side becomes: log y + x·(1/y)·dy/dx. Right side becomes: (dy/dx)·log x + y·(1/x).
5. Bring every dy/dx term to one side: (dy/dx)·(x/y − log x) = y/x − log y.
6. Multiply throughout by xy to clear the fractions, then solve for dy/dx.

Answerdy/dx = y(y − x·log y) ÷ [x(x − y·log x)]

Watch this explained “Collecting the derivative”, 20:08 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. Take log of both sides: y·log(cos x) = x·log(cos y).
2. Differentiate the left side with respect to x (product rule): (dy/dx)·log(cos x) + y·(−sin x/cos x) = (dy/dx)·log(cos x) − y·tan x.
3. Differentiate the right side (product rule, and the chain rule on log(cos y)): log(cos y) + x·(−sin y/cos y)·(dy/dx) = log(cos y) − x·tan y·(dy/dx).
4. So (dy/dx)·log(cos x) − y·tan x = log(cos y) − x·tan y·(dy/dx).
5. Bring the dy/dx terms together: (dy/dx)·[log(cos x) + x·tan y] = log(cos y) + y·tan x.
6. Divide by the bracket to get dy/dx.

Answerdy/dx = [log(cos y) + y·tan x] ÷ [log(cos x) + x·tan y]

Watch this explained “Collecting the derivative”, 20:08 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. Take the natural log of both sides: log x + log y = x − y.
2. Differentiate both sides with respect to x: 1/x + (1/y)(dy/dx) = 1 − dy/dx.
3. Collect the dy/dx terms: (dy/dx)·(1/y + 1) = 1 − 1/x.
4. Multiply throughout by y and simplify, then solve for dy/dx.

Answerdy/dx = y(x − 1) ÷ [x(1 + y)]

Watch this explained “Collecting the derivative”, 20:08 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Matches NCERT’s answer

1. Take log of both sides: log f(x) = log(1+x) + log(1+x²) + log(1+x⁴) + log(1+x⁸).
2. Differentiate each term on the right — each one becomes (derivative of the bracket) ÷ (the bracket itself).
3. So (1/f)·f′(x) = 1/(1+x) + 2x/(1+x²) + 4x³/(1+x⁴) + 8x⁷/(1+x⁸).
4. Multiply both sides by f(x) to get f′(x).
5. At x = 1: f(1) = 2×2×2×2 = 16, and the four fractions become 1/2, 1, 2 and 4, which add to 7.5.
6. f′(1) = 16 × 7.5 = 120.

Answerf′(x) = f(x)·[1/(1+x) + 2x/(1+x²) + 4x³/(1+x⁴) + 8x⁷/(1+x⁸)], and f′(1) = 120.

Watch this explained “A hundred and twenty”, 17:43 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134Checked by computer

(i) by using product rule

1. Let u = x² − 5x + 8 and v = x³ + 7x + 9.
2. Product rule: d/dx(uv) = u′v + uv′, where u′ = 2x−5 and v′ = 3x²+7.
3. So the derivative is (2x−5)(x³+7x+9) + (x²−5x+8)(3x²+7).

Answer(2x−5)(x³+7x+9) + (x²−5x+8)(3x²+7), which simplifies to 5x⁴ − 20x³ + 45x² − 52x + 11.

(ii) by expanding the product to obtain a single polynomial

1. Multiply out (x²−5x+8)(x³+7x+9) term by term and collect like powers of x.
2. This gives the single polynomial x⁵ − 5x⁴ + 15x³ − 26x² + 11x + 72.
3. Differentiate this polynomial term by term.

AnswerThe expanded polynomial is x⁵ − 5x⁴ + 15x³ − 26x² + 11x + 72, and its derivative is 5x⁴ − 20x³ + 45x² − 52x + 11.

(iii) by logarithmic differentiation

1. Let y = (x²−5x+8)(x³+7x+9). Take log: log y = log(x²−5x+8) + log(x³+7x+9).
2. Differentiate: (1/y)(dy/dx) = (2x−5)/(x²−5x+8) + (3x²+7)/(x³+7x+9).
3. Multiply both sides by y and substitute y back in, then simplify.

Answerdy/dx simplifies to 5x⁴ − 20x³ + 45x² − 52x + 11 — the same polynomial as in (i) and (ii), so yes, all three methods agree.

Watch this explained “The second use”, 12:53 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 134One way to think about it

1. Method 1 (repeated product rule): write u·v·w as u·(v·w) and apply the product rule once: d/dx[u·(vw)] = (du/dx)·(vw) + u·d/dx(vw).
2. Apply the product rule again to vw: d/dx(vw) = (dv/dx)·w + v·(dw/dx).
3. Substitute back: d/dx(uvw) = (du/dx)·v·w + u·(dv/dx)·w + u·v·(dw/dx).
4. Method 2 (logarithmic differentiation), where u, v and w are not zero: let y = uvw. Take log: log|y| = log|u| + log|v| + log|w|. (The absolute values only keep the logs defined; the derivative of log|u| is still (1/u)·du/dx.)
5. Differentiate both sides with respect to x: (1/y)(dy/dx) = (1/u)(du/dx) + (1/v)(dv/dx) + (1/w)(dw/dx).
6. Multiply both sides by y = uvw: dy/dx = (du/dx)·v·w + u·(dv/dx)·w + u·v·(dw/dx).
7. Both methods give the same expression, which is what we had to show.

In shortd/dx(u·v·w) = (du/dx)·v·w + u·(dv/dx)·w + u·v·(dw/dx), shown both by repeated product rule and by logarithmic differentiation.

Watch this explained “The second use”, 12:53 into Taking logarithms first, when the base and the power both vary