12th NCERT Continuity and Differentiability Miscellaneous 5 Questions 23

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Differentiate w.r.t x the function in Exercise 1 to 11

Question (1)

(3x² - 9x + 5 )⁹

Solution

Differentiating w.r.t. x, \[\frac{{dy}}{{dx}} = 9{\left( {3{x^2} - 9x + 5} \right)^8}\frac{d}{{dx}}\left( {3{x^2} - 9x + 5} \right)\] \[\frac{{dy}}{{dx}} = 9{\left( {3{x^2} - 9x + 5} \right)^8}\left( {6x - 9} \right)\] \[\frac{{dy}}{{dx}} = 9{\left( {3{x^2} - 9x + 5} \right)^8}3\left( {2x - 3} \right)\] \[\frac{{dy}}{{dx}} = 27\left( {2x - 3} \right){\left( {3{x^2} - 9x + 5} \right)^8}\]

Question (2)

sin³ + cos⁶x

Solution

Differentiating w.r.t. x, \[\frac{{dy}}{{dx}} = 3{\sin ^2}x\cos x + 6{\cos ^5}x \cdot \left( { - \sin x} \right)\] \[\frac{{dy}}{{dx}} = 3{\sin ^2}x\cos x - 6\sin x{\cos ^5}x\] \[\frac{{dy}}{{dx}} = 3\sin x\cos x\left( {\sin x - 2{{\cos }^4}x} \right)\]

Question (3)

(5x)^3cos2x

Solution

Take log on both side log y = log[(5x)^3cos2x]
log y = 3cos2x log(5x)
Differentiating w.r.t. x, \[\frac{1}{y}\frac{{dy}}{{dx}} = 3\left[ {\cos 2x\frac{d}{{dx}}\log 5x + \log 5x\frac{d}{{dx}}\cos 2x} \right]\] \[\frac{{dy}}{{dx}} = 3y\left[ {\frac{{\cos 2x}}{{\require{cancel} \cancel{5}x}} \times \cancel{5} + \log 5x \cdot \left( { - \sin 2x} \right)2} \right]\] \[\frac{{dy}}{{dx}} = 3{\left( {5x} \right)^{3\cos 2x}}\left[ {\frac{{\cos 2x}}{x} - 2\sin 2x\log 5x} \right]\]

Question (4)

sin^-1(x√x). 0 ≤ x ≤ 1

Solution

y = sin^-1(x^3/2)
Differentiating w.r.t. x, \[\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {{\left( {{x^{\frac{3}{2}}}} \right)}^2}} }}\frac{d}{{dx}}{x^{\frac{3}{2}}}\] \[\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {x^3}} }}\frac{3}{2}{x^{\frac{1}{2}}}\] \[\frac{{dy}}{{dx}} = \frac{3}{2}\frac{{\sqrt x }}{{\sqrt {1 - {x^3}} }}\] \[\frac{{dy}}{{dx}} = \frac{3}{2}\sqrt {\frac{x}{{1 - {x^3}}}} \]

Question (5)

\[\frac{{{{\cos }^{ - 1}}\frac{x}{2}}}{{\sqrt {2x + 7} }}\] -2 < x < 2

Solution

Differentiating w.r.t. x, \[\frac{{dy}}{{dx}} = \frac{{\sqrt {2x + 7} \frac{d}{{dx}}{{\cos }^{ - 1}}\frac{x}{2} - {{\cos }^{ - 1}}\frac{x}{2}\frac{d}{{dx}}\sqrt {2x + 7} }}{{{{\left( {\sqrt {2x + 7} } \right)}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{1}{{2x + 7}}\left[ {\sqrt {2x + 7} \times \frac{{ - 1}}{{\sqrt {1 - {{\left( {\frac{x}{2}} \right)}^2}} }} \times \frac{1}{2} - {{\cos }^{ - 1}}\left( {\frac{x}{2}} \right)\frac{1}{{\cancel{2}\sqrt {2x + 7} }} \times\cancel{2}} \right]\] \[\frac{{dy}}{{dx}} = \frac{1}{{2x + 7}}\left[ {\frac{{ - \sqrt {2x + 7} }}{{\cancel{2}\frac{{\sqrt {4 - {x^2}} }}{\cancel{2}}}} - \frac{{{{\cos }^{ - 1}}\left( {\frac{x}{2}} \right)}}{{\sqrt {2x + 7} }}} \right]\] \[\frac{{dy}}{{dx}} = \frac{{ - \sqrt {2x + 7} }}{{\left( {2x + 7} \right)\sqrt {4 - {x^2}} }} - \frac{{{{\cos }^{ - 1}}\frac{x}{2}}}{{{{\left( {2x + 7} \right)}^{\frac{3}{2}}}}}\] \[\frac{{dy}}{{dx}} = \frac{{ - 1}}{{\sqrt {2x + 7} \sqrt {4 - {x^2}} }} - \frac{{{{\cos }^{ - 1}}\frac{x}{2}}}{{{{\left( {2x + 7} \right)}^{\frac{3}{2}}}}}\]

Question (6)

\[{\cot ^{ - 1}}\left[ {\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right]\] 0 < x < (π/2)

Solution

\[y = {\cot ^{ - 1}}\left[ {\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right]\] \[y = {\cot ^{ - 1}}\left[ {\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }} \times \frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}} \right]\] \[y = {\cot ^{ - 1}}\left[ {\frac{{1 + \sin x + 1 - \sin x + 2\sqrt {1 - {{\sin }^2}x} }}{{\left( {1 + \sin x} \right) - \left( {1 - \sin x} \right)}}} \right]\] \[y = {\cot ^{ - 1}}\left[ {\frac{{2 + 2\cos x}}{{2\sin x}}} \right]\] \[y = {\cot ^{ - 1}}\left[ {\frac{{\cancel{2}\left( {1 + \cos x} \right)}}{{\cancel{2}\sin x}}} \right]\] \[y = {\cot ^{ - 1}}\left[ {\frac{{\cancel{2}{{\cos }^2}\frac{x}{2}}}{{\cancel{2}\sin \frac{x}{2}\cos \frac{x}{2}}}} \right]\] \[y = {\cot ^{ - 1}}\left( {\cot \frac{x}{2}} \right)\] \[y = \frac{x}{2}\] Differentiating w.r.t. x, \[\frac{{dy}}{{dx}} = \frac{1}{2}\]

Question (7)

(logx)^logx, x > 1

Solution

Take log on both sides
logy = log[(logx)^logx]
log y = logx [log (logx) ]
Differentiating w.r.t. x, \[\frac{1}{y}\frac{{dy}}{{dx}} = \log x\frac{d}{{dx}}\log \left( {\log x} \right) + \log \left( {\log x} \right)\frac{d}{{dx}}\log x\] \[\frac{1}{y}\frac{{dy}}{{dx}} = \log x\frac{1}{{\log x}}\frac{1}{x} + \log \left( {\log x} \right)\frac{1}{x}\] \[\frac{{dy}}{{dx}} = y\left[ {\frac{1}{x} + \frac{{\log \left( {\log x} \right)}}{x}} \right]\] \[\frac{{dy}}{{dx}} = {\left( {\log x} \right)^{\log x}}\left[ {\frac{1}{x} + \frac{{\log \left( {\log x} \right)}}{x}} \right]\]

Question (8)

cos(a cos x + b sin x), for some constant a and b

Solution

y = cos(a cos x + b sin x)
Differentiating w.r.t. x, \[\frac{{dy}}{{dx}} = - \sin \left( {a\cos x + b\sin x} \right) \cdot \frac{d}{{dx}}\left( {a\cos x + b\sin x} \right)\] \[\frac{{dy}}{{dx}} = - \sin \left( {a\cos x + b\sin x} \right) \cdot \left[ {a\left( { - \sin x} \right) + b\cos x} \right]\] \[\frac{{dy}}{{dx}} = - \sin \left( {a\cos x + b\sin x} \right)\left( { - a\sin x + b\cos x} \right)\] \[\frac{{dy}}{{dx}} = \left( {a\sin x - b\cos x} \right)\sin \left( {a\cos x + b\sin x} \right)\]

Question (9)

( sin x - cos x)^{(sinx - cosx)}, π/4 < x < 3π/4

Solution

y = ( sin x - cos x)^{(sinx - cosx)}
Take log on both side
log y = (sinx-cosx) log(sinx-cosx)
Differentiating w.r.t. x, \[\frac{1}{y}\frac{{dy}}{{dx}} = \left( {\sin x - \cos x} \right)\frac{d}{{dx}}\log \left( {\sin x - \cos x} \right) + \log \left( {\sin x - \cos x} \right)\frac{d}{{dx}}\left( {\sin x - \cos x} \right)\] \[\frac{1}{y}\frac{{dy}}{{dx}} = \frac{{\left( {\sin x - \cos x} \right)}}{{\left( {\sin x - \cos x} \right)}}\frac{d}{{dx}}\left( {\sin x - \cos x} \right) + \log \left( {\sin x - \cos x} \right)\left( {\cos x - \left( { - \sin x} \right)} \right)\] \[\frac{1}{y}\frac{{dy}}{{dx}} = \left[ {\cos x - \left( { - \sin x} \right)} \right] + \log \left( {\sin x - \cos x} \right)\left( {\cos x + \sin x} \right)\] \[\frac{{dy}}{{dx}} = y\left[ {\cos x + \sin x} \right]\left[ {1 + \log \left( {\sin x - \cos x} \right)} \right]\] \[\frac{{dy}}{{dx}} = {\left( {\sin x - \cos x} \right)^{\sin x - \cos x}}\left( {\cos x + \sin x} \right)\left[ {1 + \log \left( {\sin x - \cos x} \right)} \right]\]

Question (10)

x^x + x^a + a^x+ a^a, for some fixed a> 0 and x> 0

Solution

y = x^x + x^a + a^x+ a^a
Let u = x^x Take log on both side
log u = xlogx
Differentiating w.r.t. x, \[\frac{1}{u}\frac{{du}}{{dx}} = x\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}x\] \[\frac{1}{u}\frac{{du}}{{dx}} = x\frac{1}{x} + \log x\] \[\frac{{du}}{{dx}} = u\left[ {1 + \log x} \right]\] \[\frac{{du}}{{dx}} = {x^x}\left( {1 + \log x} \right)\] Now y = x^x + x^a + a^x+ a^a

y = u+x^a+a^x+a^a
Differentiating w.r.t. x, \[\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} + a{x^{a - 1}} + {a^x}\log a + 0\] \[\frac{{dy}}{{dx}} = {x^x}\left( {1 + \log x} \right) + a{x^{a - 1}} + {a^x}\log a\]

Question (11)

x^x2-3 + (x-3)^x2 , for x > 3

Solution

Let u = x^x2-3
Take log on both side
log u = (x²-3)log x
Differentiating w.r.t. x,
\[\frac{1}{u}\frac{{du}}{{dx}} = \left( {{x^2} - 3} \right)\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}\left( {{x^2} - 3} \right)\] \[\frac{1}{u}\frac{{du}}{{dx}} = \left( {{x^2} - 3} \right)\frac{1}{x} + \log x\left( {2x} \right)\] \[\frac{{du}}{{dx}} = u\left[ {x - \frac{3}{x} + 2x\log x} \right]\] \[\frac{{du}}{{dx}} = {x^{{x^2} - 3}}\left[ {x - \frac{3}{x} + 2x\log x} \right]\] Let v= (x-3)^x2
Take log on both side
log v = x² log(x-3)
Differentiating w.r.t. x,
\[\frac{1}{v}\frac{{dv}}{{dx}} = {x^2}\frac{d}{{dx}}\log \left( {x - 3} \right) + \log \left( {x - 3} \right)\frac{d}{{dx}}{x^2}\] \[\frac{1}{v}\frac{{dv}}{{dx}} = \frac{{{x^2}}}{{x - 3}} + 2x\log \left( {x - 3} \right)\] \[\frac{{dv}}{{dx}} = v\left[ {\frac{{{x^2}}}{{x - 3}} + 2x\log \left( {x - 3} \right)} \right]\] \[\frac{{dv}}{{dx}} = {\left( {x - 3} \right)^{{x^2}}}\left[ {\frac{{{x^2}}}{{x - 3}} + 2x\log \left( {x - 3} \right)} \right]\] Now y = (x)^x2-3 + (x-3)^x2
y = u + v
Differentiating w.r.t. x, \[\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}}\] \[\frac{{dy}}{{dx}} = {x^{{x^2} - 3}}\left[ {\frac{{{x^2} - 3}}{x} + 2x\log x} \right] + {\left( x \right)^{{x^2} - 3}}\left[ {\frac{{{x^2}}}{{x - 3}} + 2x\log \left( {x - 3} \right)} \right]\]

Question (12)

Find dy/dx. if y =12 ( 1 - cost),
x = 10(t - sin t) , π/2 < t < π/2

Solution

Differentiating w.r.t. t, \[\frac{{dy}}{{dt}} = 12\left( {0 - \left( { - \sin t} \right)} \right)\] \[\frac{{dy}}{{dt}} = 12\sin t\] x = 10 (t - sint)
Differentiating w.r.t. t, \[\frac{{dx}}{{dt}} = 10\left( {1 - \cos t} \right)\] \[\text{Now} \quad \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{12\sin t}}{{10\left( {1 - \cos t} \right)}}\] \[\frac{{dy}}{{dx}} = \frac{{\cancel{12}^6\left( {\cancel{2}\sin \frac{t}{2}\cos \frac{t}{2}} \right)}}{{\cancel{10}_5\left( {\cancel{2}{{\sin }^2}\frac{t}{2}} \right)}}\] \[\frac{{dy}}{{dx}} = \frac{6}{5}\frac{{\cos \frac{t}{2}}}{{\sin \frac{t}{2}}}\] \[\frac{{dy}}{{dx}} = \frac{6}{5}\cot \left( {\frac{t}{2}} \right)\]

Question (13)

Find dy/dx. if y = sin^-1x + sin^-1 √(1-x²), -1≤ x ≤ 1

Solution

\[y = {\sin ^{ - 1}}x + {\sin ^{ - 1}}\sqrt {1 - {x^2}} \] Differentiating w.r.t. x, \[\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }} + \frac{1}{{\sqrt {1 - {{\left( {\sqrt {1 - {x^2}} } \right)}^2}} }}\frac{d}{{dx}}\sqrt {1 - {x^2}} \] \[\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }} + \frac{1}{{\sqrt {1 - 1 + {x^2}} }}\frac{1}{{2\sqrt {1 - {x^2}} }}\left( { - 2x} \right)\] \[\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }} + \frac{1}{x}\frac{{ - x}}{{\sqrt {1 - {x^2}} }}\] \[\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{\sqrt {1 - {x^2}} }} = 0\]

Question (14)

\[x\sqrt {1 + y} + y\sqrt {1 + x} = 0\] for -1 < x < 1, prove that \[\frac{{dy}}{{dx}} = - \frac{1}{{{{\left( {1 + x} \right)}^2}}}\]

Solution

\[x\sqrt {1 + y} = - y\sqrt {1 + x} \] Squaring on both side
x² ( 1 + y) = y² ( 1 + x)
x² + x²y = y² + xy²
x²y - xy²+x²-y² = 0
xy(x-y) + (x-y)(x+y) = 0
(x-y)(xy+x+y) = 0
x - y = 0  OR  xy+x+y = 0
y = x  OR  xy + y= -x
y=x  OR  y = -x/1+x
but if y = x it will not satisfy the equation \[\therefore y = \frac{{ - x}}{{1 + x}}\] Differentiating w.r.t. x, \[\frac{{dy}}{{dx}} = - \left[ {\frac{{\left( {1 + x} \right)\frac{d}{{dx}}x - x\frac{d}{{dx}}\left( {1 + x} \right)}}{{{{\left( {1 + x} \right)}^2}}}} \right]\] \[\frac{{dy}}{{dx}} = - \left[ {\frac{{\left( {1 + x} \right)1 - x}}{{{{\left( {1 + x} \right)}^2}}}} \right]\] \[\frac{{dy}}{{dx}} = - \left[ {\frac{{1 + x - x}}{{{{\left( {1 + x} \right)}^2}}}} \right]\] \[\frac{{dy}}{{dx}} = - \frac{1}{{{{\left( {1 + x} \right)}^2}}}\]

Question (15)

If (x-a)² + (y-b)² = c², for some c > 0, prove that \[\frac{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}}\] is a constant independant of a and b

Solution

Differentiating w.r.t. x, \[2\left( {x - a} \right) + 2\left( {y - b} \right)\frac{{dy}}{{dx}} = 0\] \[2\left( {y - b} \right)\frac{{dy}}{{dx}} = - 2\left( {x - a} \right)\] \[\frac{{dy}}{{dx}} = \frac{{ - \left( {x - a} \right)}}{{\left( {y - b} \right)}}\] Differentiating w.r.t. x, again \[\frac{{{d^2}y}}{{d{x^2}}} = - \left[ {\frac{{\left( {y - b} \right)\frac{d}{{dx}}\left( {x - a} \right) - \left( {x - a} \right)\frac{d}{{dx}}\left( {y - b} \right)}}{{{{\left( {y - b} \right)}^2}}}} \right]\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 1}}{{{{\left( {y - b} \right)}^2}}}\left[ {\left( {y - b} \right) - \left( {x - a} \right)\frac{d}{{dx}}y} \right]\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 1}}{{{{\left( {y - b} \right)}^2}}}\left[ {\left( {y - b} \right) - \left( {x - a} \right)\left( {\frac{{ - \left( {x - a} \right)}}{{y - b}}} \right)} \right]\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 1}}{{{{\left( {y - b} \right)}^2}}}\left[ {\left( {y - b} \right) + \frac{{{{\left( {x - a} \right)}^2}}}{{y - b}}} \right]\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 1}}{{{{\left( {y - b} \right)}^2}}}\left[ {\frac{{{{\left( {y - b} \right)}^2} + {{\left( {x - a} \right)}^2}}}{{y - b}}} \right]\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - {c^2}}}{{{{\left( {y - b} \right)}^3}}}\] \[LHS = \frac{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}}\] \[LHS = \frac{{{{\left[ {1 + {{\left( {\frac{{ - \left( {x - a} \right)}}{{y - b}}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{ - {c^2}}}{{{{\left( {y - b} \right)}^3}}}}}\] \[LHS = \frac{{{{\left( {y - b} \right)}^3}}}{{-{c^2}}} \times {\left[ {1 + \frac{{{{\left( {x - a} \right)}^2}}}{{{{\left( {y - b} \right)}^2}}}} \right]^{\frac{3}{2}}}\] \[LHS = \frac{{{{\left( {y - b} \right)}^3}}}{{-{c^2}}} \times {\left[ {\frac{{{{\left( {y - b} \right)}^2} + {{\left( {x - a} \right)}^2}}}{{{{\left( {y - b} \right)}^2}}}} \right]^{\frac{3}{2}}}\] \[LHS = \frac{{{{\left( {y - b} \right)}^3}}}{{ - {c^2}}} \times {\left( {\frac{{{c^2}}}{{{{\left( {y - b} \right)}^2}}}} \right)^{\frac{3}{2}}}\] \[LHS = \frac{{{{\left( {y - b} \right)}^3}}}{{ - {c^2}}} \times \frac{{{c^3}}}{{{{\left( {y - b} \right)}^3}}}\] \[LHS = - c\] LHS = constant

Question (16)

If cos y = x cos (a+y), with cos a ≠ ±1 prove that \[\frac{{dy}}{{dx}} = \frac{{{{\cos }^2}\left( {a + y} \right)}}{{\sin a}}\]

Solution

\[\therefore x = \frac{{\cos y}}{{\cos \left( {a + y} \right)}}\] Differentiating w.r.t. y, \[\frac{{dx}}{{dy}} = \frac{{\cos \left( {a + y} \right)\frac{d}{{dy}}\cos y - \cos y\frac{d}{{dy}}\cos \left( {a + y} \right)}}{{{{\cos }^2}\left( {a + y} \right)}}\] \[\frac{{dx}}{{dy}} = \frac{{\cos \left( {a + y} \right)\left( { - \sin y} \right) - \cos y\left( { - \sin \left( {a + y} \right)} \right)}}{{{{\cos }^2}\left( {a + y} \right)}}\] \[\frac{{dx}}{{dy}} = \frac{{ - \cos \left( {a + y} \right)\sin y + \sin \left( {a + y} \right)\cos y}}{{{{\cos }^2}\left( {a + y} \right)}}\] \[\frac{{dx}}{{dy}} = \frac{{\sin \left( {a + y} \right)\cos y - \cos \left( {a + y} \right)\sin y}}{{{{\cos }^2}\left( {a + y} \right)}}\] \[\frac{{dy}}{{dx}} = \frac{{\sin \left( {a + y - y} \right)}}{{{{\cos }^2}\left( {a + y} \right)}}\] \[\frac{{dx}}{{dy}} = \frac{{\sin a}}{{{{\cos }^2}\left( {a + y} \right)}}\] \[\frac{{dy}}{{dx}} = \frac{1}{{\frac{{dx}}{{dy}}}}\] \[\frac{{dy}}{{dx}} = \frac{{{{\cos }^2}\left( {a + y} \right)}}{{\sin a}}\]

Question (17)

If x = a(cost + t sint) and y= a(sint - tcost) find \[\frac{{{d^2}y}}{{d{x^2}}}\]

Solution

Differentiating w.r.t. t \[\frac{{dx}}{{dt}} = a\left[ { - \sin t + \left( {t\frac{d}{{dt}}\sin t + \sin t\frac{d}{{dt}}t} \right)} \right]\] \[\frac{{dx}}{{dt}} = a\left[ { - \sin t + t\cos t + \sin t} \right]\] \[\frac{{dx}}{{dt}} = at\cos t\] Differentiating w.r.t. t \[\frac{{dy}}{{dt}} = a\left[ {\cos t - \left( {t\frac{d}{{dt}}\cos t + \cos t\frac{d}{{dt}}t} \right)} \right]\] \[\frac{{dy}}{{dt}} = a\left[ {\cos t - t\left( { - \sin t} \right) - \cos t} \right]\] \[\frac{{dy}}{{dt}} = at\sin t\] \[\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\] \[\frac{{dy}}{{dx}} = \frac{{at\sin t}}{{at\cos t}}\] \[\frac{{dy}}{{dx}} = \tan t\] Differentiating w.r.t. x \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\tan t\] \[\frac{{{d^2}y}}{{d{x^2}}} = {\sec ^2}t\frac{{dt}}{{dx}}\] \[\frac{{{d^2}y}}{{d{x^2}}} = {\sec ^2}t\frac{1}{{at\cos t}}\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{{{{\sec }^3}t}}{{at}}\]

Question (18)

If f(x) = |x|³, show that f"(x) exists for all real x and find it

Solution

Let f(x) = x³
If x ≥ 0
f(x) = x³
f'(x) = 3x²
f"(x) = 6x
If x< 0
f(x) = - x³
f'(x) = -3x²
f"(x) = -6x
f"(x) exists for all real number x
f"(x) = 6x , x ≥ 0
f"(x) = -6x, x< 0

Question (19)

Using mathematical induction prove that \[\frac{d}{{dx}}{x^n} = n{x^{n - 1}}\] for all positive inegers n

Solution

Let us prove for n = 1
\[\text{LHS} =\qquad \frac{d}{{dx}}x = 1\] RHS= 1x^1-1 = 1 = LHS
It is true for n=1
Let us assume it is true for n = k \[\therefore \frac{d}{{dx}}{x^k} = k{x^{k - 1}}\] Let us prove for n = k+1
that is prove \[\frac{d}{{dx}}{x^{k + 1}} = \left( {k + 1} \right){x^k}\] Now \[\frac{d}{{dx}}{x^{k + 1}} = \frac{d}{{dx}}\left( {x \cdot {x^k}} \right)\] by product rule \[\frac{d}{{dx}}{x^{k + 1}} = x\frac{d}{{dx}}{x^k} + {x^k}\frac{d}{{dx}}x\] \[\frac{d}{{dx}}{x^{k + 1}} = x \cdot k{x^{k - 1}} + {x^k}\left( 1 \right)\] \[\frac{d}{{dx}}{x^{k + 1}} = k{x^{k - 1 + 1}} + {x^k}\] \[\frac{d}{{dx}}{x^{k + 1}} = k{x^k} + {x^k}\] \[\frac{d}{{dx}}{x^{k + 1}} = {x^k}\left( {k + 1} \right)\] = RHS
∴ p(n) is true for x = k+1
∴ PMI. \[ \frac{d}{{dx}}{x^n} = n{x^{n - 1}} \qquad \text{true for all n ∈N}\]

Question (20)

Using the fact that sin(A+B) = sinAcosB + cosAsinB and the differentiation obtain the sum formula for cosines

Solution

Differentiating w.r.t. x \[\cos \left( {A + B} \right)\frac{d}{{dx}}\left( {A + B} \right) = \sin A\frac{d}{{dx}}\cos B + \cos B\frac{d}{{dx}}\sin A + \cos A\frac{d}{{dx}}\sin B + \sin B\frac{d}{{dx}}\cos A\] \[\cos \left( {A + B} \right)\left( {\frac{{dA}}{{dx}} + \frac{{dB}}{{dx}}} \right) = \sin A\left( { - \sin B} \right)\frac{{dB}}{{dx}} + \cos B\cos A\frac{{dA}}{{dx}} + \cos A\cos B\frac{{dB}}{{dx}} + \sin B\left( { - \sin A} \right)\frac{{dA}}{{dx}}\] \[\cos \left( {A + B} \right)\frac{d}{{dx}}\left( {A + B} \right) = \left( {\cos A\cos B - \sin A\sin B} \right)\frac{{dB}}{{dx}} + \left( {\cos A\cos B - \sin A\sin B} \right)\frac{{dA}}{{dx}}\] \[\cos \left( {A + B} \right)\left( {\frac{{dA}}{{dx}} + \frac{{dB}}{{dx}}} \right) = \left( {\cos A\cos B - \sin A\sin B} \right)\left( {\frac{{dA}}{{dx}} + \frac{{dB}}{{dx}}} \right)\] \[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\]

Question (21)

Does there exist a function which is continuous everywhere but not differentiatiable at exactly two points? Justify your answer.

Solution

Yes it exists that function is continuous at every value of x, but not differentiable at two points
To justfy we will consider function f as follows.
f(x) = |x| - |x + 1| \[\begin{array}{l}if\;x \le - 1,\\|x + 1| = - (x + 1),|x| = - x\end{array}\] \[f(x) = - x - [ - (x + 1)] = 1\] \[\begin{array}{l}if\; - 1 < x \le 0,\\|x + 1| = x + 1,|x| = - x\end{array}\] \[f(x) = - x - (x + 1) = - 2x - 1\] \[\begin{array}{l}if\;x > 0,\\|x| = x,|x + 1| = x + 1\end{array}\] \[f(x) = x - (x + 1) = - 1\] \[So\;\;f(x) = 1\;\;if\;x \le - 1\] \[ = - 2x - 1\;if\; - 1 < x \le 0\] \[ = - 1\;if\;x > 0\] The function is break at x = -1, and x = 0.so check the continuity at x = -1 and at x = 0. For the other value of x it is continuous.
Let us check at x = -1, f(-1) = -2(-1)-1 = 2 - 1 = 1
\[LHL = \mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ - }} 1 = 1\] \[RHL = \mathop {\lim }\limits_{x \to - {1^ + }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} - 2x - 1 = - 2( - 1) - 1 = 1\] \[\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = f( - 1) = \mathop {\lim }\limits_{x \to - {1^ + }} f(x)\] So f(x) is continuous at x = -1.
Let us check at x = 0, f(0) = -2(0) - 1 = 0 - 1 = - 1.
\[LHL = \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to - {0^ + }} - 2x - 1 = - 2(0) - 1 = - 1\] \[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to - {0^ + }} - 1 = -1\] \[\mathop {\lim }\limits_{x \to - {0^ - }} f(x) = f(0) = \mathop {\lim }\limits_{x \to - {0^ + }} f(x)\] So f(x) is continuous at x = 0.
So f(x) is continuous at every point of x. Let us check the differeniability at x = - 1
f(-1) = 1
\[lhl = \mathop {\lim }\limits_{x \to - {1^ - }} \frac{{f(x) - f( - 1)}}{{x - ( - 1)}} = \mathop {\lim }\limits_{x \to - {1^ - }} \frac{{1 - 1}}{{x + 1}} = 0\] \[Rhl = \mathop {\lim }\limits_{x \to - {1^ + }} \frac{{f(x) - f( - 1)}}{{x - ( - 1)}} = \mathop {\lim }\limits_{x \to - {1^ + }} \frac{{ - 2x - 1 - 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to - {1^ + }} \frac{{ - 2(x + 1)}}{{x + 1}} = - 2\] \[\mathop {\lim }\limits_{x \to - {1^ - }} \frac{{f(x) - f( - 1)}}{{x - ( - 1)}} \ne \mathop {\lim }\limits_{x \to - {1^ + }} \frac{{f(x) - f( - 1)}}{{x - ( - 1)}}\] so f is not differentiable at x= - 1.
To check at x = 0
f(0) = -2(0) - 1 = -1.
\[\mathop {\lim }\limits_{x \to {0^ - }} \frac{{f(x) - f(0)}}{{x - (0)}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - 2x - 1 - ( - 1)}}{x} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - 2(x)}}{x} = - 2\] \[\mathop {\lim }\limits_{x \to {0^ + }} \frac{{f(x) - f(0)}}{{x - (0)}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{ - 1 - ( - 1)}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{0}{x} = 0\] \[\mathop {\lim }\limits_{x \to {0^ - }} \frac{{f(x) - f(0)}}{{x - (0)}} \ne \mathop {\lim }\limits_{x \to {0^ + }} \frac{{f(x) - f(0)}}{{x - (0)}}\] Hence f(x) is also not differentiable at x = 0.
Hence the function was continuous at every point but not differentiable at two points.

Question (22)

\[\text{If}\qquad y = \left| {\begin{array}{*{20}{c}}{f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)}\\l&m&n\\a&b&c\end{array}} \right|\] Prove that \[\frac{{dy}}{{dx}} = \left| {\begin{array}{*{20}{c}}{f'\left( x \right)}&{g'\left( x \right)}&{h'\left( x \right)}\\l&m&n\\a& b&c\end{array}} \right|\]

Solution

\[\text{If}\qquad y = \left| {\begin{array}{*{20}{c}}{f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)}\\l&m&n\\a&b&c\end{array}} \right|\] Expanding this matrics along first row we get,
y=f(x)[mc - bn] - g(x) [ lc - an ] + h(x) [ lb - ma]
Diff. w.r. t. x, we get ,
dy / dx = (mc - bn)f'(x) - ( lc - an)g'(x) + ( lb - ma) h'(x)
it can be rearrange in the matrix as
\[\frac{{dy}}{{dx}} = \left| {\begin{array}{*{20}{c}}{f'\left( x \right)}&{g'\left( x \right)}&{h'\left( x \right)}\\l&m&n\\a& b&c\end{array}} \right|\]

Question (23)

If y = e^acos-1x , -1 ≤ x ≤ 1, show that \[\left( {1 - {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}} - x\frac{{dy}}{{dx}} - {a^2}y = 0\]

Solution

Differentiating w.r.t. x \[\frac{{dy}}{{dx}} = {e^{a{{\cos }^{ - 1}}x}}\frac{d}{{dx}}\left( {a{{\cos }^{ - 1}}x} \right)\] \[{y_1} = {e^{a{{\cos }^{ - 1}}x}}\frac{{ - a}}{{\sqrt {1 - {x^2}} }}\] \[\sqrt {1 - {x^2}} {y_1} = - a{e^{a{{\cos }^{ - 1}}x}}\] Differentiating w.r.t. x again \[\sqrt {1 - {x^2}} \frac{{d{y_1}}}{{dx}} + {y_1}\frac{d}{{dx}}\sqrt {1 - {x^2}} = - a{e^{a{{\cos }^{ - 1}}x}} \cdot \frac{d}{{dx}}\left( {a{{\cos }^{ - 1}}x} \right)\] \[\sqrt {1 - {x^2}} {y_2} + {y_1}\left( {\frac{{ - 2x}}{{2\sqrt {1 - {x^2}} }}} \right) = - a{e^{a{{\cos }^{ - 1}}x}} \cdot \frac{{ - a}}{{\sqrt {1 - {x^2}} }}\] \[\sqrt {1 - {x^2}} {y_2} - \frac{{x{y_1}}}{{\sqrt {1 - {x^2}} }} = \frac{{{a^2}{e^{a{{\cos }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }}\] (1-x²)y₂ - xy₁ = a²y \[\left( {1 - {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}} - x\frac{{dy}}{{dx}} - {a^2}y = 0\]