2<\/sup>t) = 3 cos t . cos 2t
\n<\/p>\n(ii) x = \\(\\frac{3 a t}{1+t^3}\\), y = \\(\\frac{3 a t^2}{1+t^3}\\) (E.Q.) [Board Model paper]
\nAnswer:
\n<\/p>\n
<\/p>\n
(iii) x = a (cos t + t sin t),
\ny = a (sin t – t cos t). (S.A.Q.)
\nAnswer:
\n\\(\\frac{d x}{d t}\\) = a(- sin t + t cos t + sin t)
\n= at cos t
\n\\(\\frac{d y}{d t}\\) = a [cos t – (- t sin t + cos t)] = at sin t
\n<\/p>\n
(iv) x = a\\(\\left[\\frac{1-t^2}{1+t^2}\\right]\\), y = \\(\\frac{2 b t}{1+t^2}\\) (S.A.Q.)
\nAnswer:
\n<\/p>\n
Question 3.
\nDifferentIate f(x) with respect to g(x) for the following. (S.A.Q.)
\n(i) f(x) = loga<\/sub>x<\/sup>, g(x) = ax<\/sup>
\nAnswer:
\nLet y = loga<\/sub>x<\/sup> and z = ax<\/sup>
\n<\/p>\n(ii) f(x) = sec-1<\/sup>\\(\\left(\\frac{1}{2 x^2-1}\\right)\\), g(x) = \\(\\sqrt{1-x^2}\\)
\nAnswer:
\n<\/p>\n<\/p>\n
(iii) f(x) = tan-1<\/sup>\\(\\left(\\frac{\\sqrt{1+x^2}-1}{x}\\right)\\), g(x) = tan-1<\/sup>x [June 2009 IPE]
\nAnswer:
\n<\/p>\nQuestion 4.
\nFind the derivative of the function y defined implicitly, by each of the following equations. (S.A.Q.)
\n(i) x4<\/sup> + y4<\/sup> – a2<\/sup> xy = 0
\nAnswer:
\nDifferentiating with respect to \u2018x
\n<\/p>\n(ii) y = xy<\/sup> (S.A.Q.) (March 2004)
\nAnswer:
\nGiven y = xy<\/sup>
\nThen log y = y log x ………………. (1)
\nDifferentiating w.r.t. to ‘x’ both sides
\n<\/p>\n<\/p>\n
(iii) yx<\/sup> = xsin y<\/sup> (S.A.Q.)
\nAnswer:
\nTaking logarithm on both sides
\nx log y = sin y . log x
\nDifferentiating w.r.t to ‘x’
\n<\/p>\nQuestion 5.
\nEstablish the following.
\n(i) If \\(\\sqrt{1-x^2}\\) + \\(\\sqrt{1-y^2}\\) = a(x – y) then \\(\\frac{d y}{d x}=\\frac{\\sqrt{1-y^2}}{\\sqrt{1-x^2}}\\) (E.Q.) (May 2014, March 2009)
\nAnswer:
\nTake x = sin \u03b8 and y = sin \u03a6
\n\u2234 From the given condition
\n<\/p>\n
(ii) If y = x\\(\\sqrt{a^2+x^2}\\) + a2<\/sup> log(x + \\(\\sqrt{a^2+x^2}\\)) then \\(\\frac{\\mathbf{d y}}{\\mathbf{d x}}\\) = 2\\(\\sqrt{a^2+x^2}\\) (E.Q.)
\nAnswer:
\n<\/p>\n<\/p>\n
(iii) If xlog y<\/sup> = log x then \\(\\frac{d y}{d x}\\) = \\(\\left[\\frac{1-\\log x \\log y}{(\\log x)^2}\\right]\\) (S.A.Q)
\nAnswer:
\nGiven xlog y<\/sup> = log x
\nTaking logarithm on both sides
\nlog y . log x = log (log x)
\nDifferentiating with respect to ‘x’
\n<\/p>\n(iv) If y = tan-1<\/sup>\\(\\left(\\frac{2 x}{1-x^2}\\right)\\) + tan-1<\/sup>\\(\\left(\\frac{3 x-x^3}{1-3 x^2}\\right)\\) – tan-1<\/sup>\\(\\left(\\frac{4 x-4 x^3}{1-6 x^2+x^4}\\right)\\) then \\(\\frac{d y}{d x}=\\frac{1}{1+x^2}\\) (S.A.Q.)
\nAnswer:
\nLet x = tan \u03b8, then
\n
\n= tan-1<\/sup>(tan 2\u03b8) + tan-1<\/sup> (tan 3\u03b8) – tan-1<\/sup> (tan 4\u03b8)
\n= 2\u03b8 + 3\u03b8 – 4\u03b8 = \u03b8 = tan-1<\/sup>x
\n\u2234 \\(\\frac{d y}{d x}=\\frac{1}{1+x^2}\\)<\/p>\n(v) If xy<\/sup> = yx<\/sup> then \\(\\) (S.A.Q.)
\nAnswer:
\nGiven xy<\/sup> = yx<\/sup>
\nThen log xy<\/sup> = log yx<\/sup>
\n\u21d2 y log x = x log y
\nDifferentiating both sides w.r.t to ‘x’
\n<\/p>\n(vi) If x\\(\\frac{2}{3}\\)<\/sup> + y\\(\\frac{2}{3}\\)<\/sup> = a\\(\\frac{2}{3}\\)<\/sup> then \\(\\frac{d y}{d x}=-\\sqrt[3]{\\frac{y}{x}}\\) (S.A.Q.)
\nAnswer:
\nGiven x\\(\\frac{2}{3}\\)<\/sup> + y\\(\\frac{2}{3}\\)<\/sup> = a\\(\\frac{2}{3}\\)<\/sup>
\nThen differentiating both sides w.r.t to ‘x’
\n<\/p>\n<\/p>\n
Question 6.
\nFind \\(\\frac{d y}{d x}\\) of each of the following functions. (S.A.Q.)<\/p>\n
(i) y = \\(\\frac{(1-2 x)^{\\frac{2}{3}}(1+3 x)^{\\frac{-3}{4}}}{(1-6 x)^{\\frac{5}{6}}(1+7 x)^{\\frac{-6}{7}}}\\)
\nAnswer:
\n<\/p>\n
(ii) y = \\(\\frac{x^4 \\sqrt[3]{x^2+4}}{\\sqrt{4 x^2-7}}\\)
\nAnswer:
\n<\/p>\n
(iii) y = \\(\\frac{(a-x)^2(b-x)^3}{(c-2 x)^3}\\)
\nAnswer:
\nlog y = log (a – x)2<\/sup> + log(b – x)3<\/sup> – log (c – 2x)3<\/sup>
\n= 2 log (a – x) + 3 log (b – x) – 3 log (c – 2x)
\nDifferentiating both sides w.r.t to ‘x’
\n<\/p>\n<\/p>\n
(iv) y = \\(\\frac{x^3 \\sqrt{2+3 x}}{(2+x)(1-x)}\\)
\nAnswer:
\nlog y = log \\(\\left[\\frac{x^3 \\sqrt{2+3 x}}{(2+x)(1-x)}\\right]\\)
\n= log x3<\/sup> + \\(\\frac{1}{2}\\) log (2 + 3x) – log (2 + x) – log (1 – x)
\n= 3 log x + \\(\\frac{1}{2}\\) log (2 + 3x) – log (2 + x) – log (1 – x)
\nDifferentiating both sides w.r.t to ‘x’
\n<\/p>\n(v) y = \\(\\sqrt{\\frac{(x-3)\\left(x^2+4\\right)}{3 x^2+4 x+5}}\\)
\nAnswer:
\n<\/p>\n
III.
\nQuestion 1.
\nFind the derivatives of the following functions. (E.Q.) (March 2013)
\n(i) (sin x)log x<\/sup> + xsin x<\/sup>
\nAnswer:
\ny = (sin x)log x<\/sup> + xsin x<\/sup>
\nLet y = u + v where u = (sin x)log x<\/sup> ………….. (1)
\nand v = xsin x<\/sup> ……………….. (2)
\nFrom (1) log u = log x log (sin x)
\nDifferentiate w.r.to x both sides
\n<\/p>\n<\/p>\n
(ii) xxx<\/sup><\/sup>
\nAnswer:
\nLet y = xxx<\/sup><\/sup>
\n\u2234 log y = log x(xx<\/sup>)<\/sup> = xx<\/sup> log x
\nDifferentiate both sides w.r.t. ‘x’.
\n\\(\\frac{1}{y} \\frac{d y}{d x}\\) = xx<\/sup> \\(\\left(\\frac{1}{x}\\right)\\) + log x . xx<\/sup> (1 + log x)
\n(\u2235 \\(\\frac{d y}{d x}\\) (xx<\/sup>) = xx<\/sup> (1 + log x))
\n= xx – 1<\/sup> + xx<\/sup> . (1 + log x). log x
\n= xx – 1<\/sup> [1 + x log x . log ex]
\n\u2234 \\(\\frac{d y}{d x}\\) = xxx<\/sup><\/sup> xx – 1<\/sup> [1 + x log x log ex]
\n= xxx<\/sup> + x – 1<\/sup> (1 + x log x log ex)<\/p>\n(iii) (sin x)x<\/sup> = xsin x<\/sup>
\nAnswer:
\nLet y = u + v where u = (sin x)x<\/sup> ………………… (1)
\nand v = xsin x<\/sup> …………………… (2)
\n<\/p>\n(iv) xx<\/sup> + (cot x)x<\/sup>
\nAnswer:
\nLet y = xx<\/sup> + (cot x)x<\/sup>
\nand suppose y = u + v and
\n\\(\\frac{d y}{d x}=\\frac{d u}{d x}+\\frac{d v}{d x}\\) ………………… (1)
\nWhere u = xx<\/sup> …………………… (2)
\nand v = (cot x)x<\/sup> …………………….. (3)
\nFrom (2)
\nu = xx<\/sup> \u21d2 \\(\\frac{\\mathrm{du}}{\\mathrm{dx}}\\) = xx<\/sup> (1 + log x)
\nFrom (3) v = (cot x)x<\/sup>
\n\u21d2 log v = x log (cot x)
\nDifferentiating both sides w.r.t to ‘x’
\n<\/p>\n<\/p>\n
Question 2.
\nEstablish the following (E.Q.)
\n(i) If xy<\/sup> + yx<\/sup> = ab<\/sup> then
\n\\(\\frac{\\mathbf{d y}}{\\mathbf{d x}}\\) = – \\(\\left(\\frac{y \\cdot x^{y-1}+y^x \\cdot \\log y}{x^y \\log x+x \\cdot y^{x-1}}\\right)\\)
\nAnswer:
\nLet u = xy<\/sup> ………………… (1) and v = yx<\/sup> …………………… (2)
\nthen given u + v = ab<\/sup> ……………………… (3)
\nFrom (1), log u = y log x
\n<\/p>\n(ii) If f(x) = sin-1<\/sup>\\(\\sqrt{\\frac{x-\\beta}{\\alpha-\\beta}}\\) and g(x) = tan-1<\/sup>\\(\\sqrt{\\frac{\\mathbf{x}-\\beta}{\\alpha-\\mathbf{x}}}\\) then f'(x) = g'(x) (\u03b2 < x < \u03b1) (March 2006)
\nAnswer:
\n<\/p>\n(iii) If a > b > 0 and 0 < x < \u03c0;
\nf(x) = (a2<\/sup> – b2<\/sup>)-1\/2<\/sup> . cos-1<\/sup>\\(\\left(\\frac{a \\cos x+b}{a+b \\cos x}\\right)\\) then f'(x) = (a + b cos x)-1<\/sup>
\nAnswer:
\n<\/p>\n<\/p>\n
Question 3.
\nDifferentiate (x2<\/sup> – 5x + 8) (x3<\/sup> + 7x + 9) by
\n(i) Using product rule
\n(ii) Obtaining a single polynomial and expanding the product
\n(iii) Logarithmic differentiation.
\nDo they all give the same answer? (EQ.)
\nAnswer:
\n(i) Using product rule:
\n\\(\\frac{d}{d x}\\) (uv) = u\\(\\frac{d v}{d x}\\) + v\\(\\frac{d u}{d x}\\)
\n\\(\\frac{d}{d x}\\)[(x2<\/sup> – 5x + 8) (x3<\/sup> + 7x + 9)
\n= (x2<\/sup> – 5x + 8) \\(\\frac{d}{d x}\\) (x3<\/sup> + 7x + 9) + (x3<\/sup> + 7x + 9) \\(\\frac{d}{d x}\\) (x2<\/sup> – 5x + 8)
\n= (x2<\/sup> – 5x + 8) (3x2<\/sup> + 7) + (x2<\/sup> + 7x + 9) (2x – 5)
\n= 3x4<\/sup> – 15x3<\/sup> + 24x2<\/sup> + 7x2<\/sup> – 35x + 56 + 2x4<\/sup> + 14x2<\/sup> + 18x – 5x3<\/sup> – 35x – 45
\n= 5x4<\/sup> – 20x3<\/sup> + 45x2<\/sup> – 52x + 11<\/p>\n(ii) Expanding the product:
\ny = (x2<\/sup> – 5x + 8) (x3<\/sup> + 7x + 9)
\n= x5<\/sup> + 7x3<\/sup> + 9x