12th NCERT Continuity and Differentiability Exercise 5.3 Questions 15

Find dy/dx of following

Question (1)

2x+3y = sinx

Solution

\[\frac{d}{{dx}}x = 1\] \[\frac{d}{{dx}}\sin x = \cos x\]
2x + 3y = sin x
differentiate w.r.t x \[2 + 3\frac{{dy}}{{dx}} = \cos x\] \[3\frac{{dy}}{{dx}} = \cos x - 2\] \[\frac{{dy}}{{dx}} = \frac{{\cos x - 2}}{3}\]

Question (2)

2x+3y = siny

Solution

\[\frac{d}{{dx}}\sin x = \cos x\] \[\frac{d}{{dx}}x = 1\]
2x + 3y = sin y
differentiate w.r.t x \[2 + 3\frac{{dy}}{{dx}} = \cos y\frac{{dy}}{{dx}}\] \[\cos y\frac{{dy}}{{dx}} - 3\frac{{dy}}{{dx}} = 2\] \[\left( {\cos y - 3} \right)\frac{{dy}}{{dx}} = 2\] \[\frac{{dy}}{{dx}} = \frac{2}{{\left( {\cos y - 3} \right)}}\]

Question (3)

ax + by2 = cosy

Solution

\[\frac{d}{{dx}}{x^n} = n{x^{n - 1}}\] \[\frac{d}{{dx}}\cos x = - \sin x\]
ax + by2 = cosy
differentiate w.r.t x \[a + 2by\frac{{dy}}{{dx}} = - \sin y\frac{{dy}}{{dx}}\] \[2by\frac{{dy}}{{dx}} + \sin y\frac{{dy}}{{dx}} = - a\] \[\frac{{dy}}{{dx}}\left( {2by + \sin y} \right) = - a\] \[\frac{{dy}}{{dx}} = \frac{{ - a}}{{\left( {2by + \sin y} \right)}}\]

Question (4)

xy + y2 = tanx + y

Solution

\[\frac{d}{{dx}}{x^n} = n{x^{n - 1}}\] \[\frac{d}{{dx}}\tan x = {\sec ^2}x\] \[\frac{d}{{dx}}\tan x = {\sec ^2}x\]
xy + y2 = tanx + y
Diff. w.r.t. x \[x\frac{d}{{dx}}y + y\frac{d}{{dx}}x + 2y\frac{{dy}}{{dx}} = {\sec ^2}x + \frac{{dy}}{{dx}}\] \[x\frac{{dy}}{{dx}} + y + 2y\frac{{dy}}{{dx}} = {\sec ^2}x + \frac{{dy}}{{dx}}\] \[\left( {x + 2y - 1} \right)\frac{{dy}}{{dx}} = {\sec ^2}x - y\] \[\frac{{dy}}{{dx}} = \frac{{{{\sec }^2}x - y}}{{x + 2y - 1}}\]

Question (5)

x2+xy+y2=100

Solution

\[\frac{d}{{dx}}uv = u\frac{{dv}}{{dx}} + v\frac{{du}}{{dx}}\] \[\frac{d}{{dx}}{x^n} = n{x^{n - 1}}\] \[\frac{d}{{dx}}\left( c \right) = 0\] c= constant
x2+xy+y2=100
Diff.w.r.t.x, \[2x + \left[ {x\frac{{dy}}{{dx}} + y\frac{d}{{dx}}x} \right] + 2y\frac{{dy}}{{dx}} = 0\] \[2x + x\frac{{dy}}{{dx}} + y + 2y\frac{{dy}}{{dx}} = 0\] \[\left( {x + 2y} \right)\frac{{dy}}{{dx}} = - \left( {2x + y} \right)\] \[\frac{{dy}}{{dx}} = \frac{{ - \left( {2x + y} \right)}}{{\left( {x + 2y} \right)}}\]

Question (6)

x3 + x2y +xy2 + y3 = 81

Solution

\[\frac{d}{{dx}}uv = u\frac{{dv}}{{dx}} + v\frac{{du}}{{dx}}\] \[\frac{d}{{dx}}{x^n} = n{x^{n - 1}}\] \[\frac{d}{{dx}}\left( c \right) = 0\] c= constant
x3 + x2y +xy2 + y3 = 81
Diff.w.r.t.x, \[3{x^2} + \left[ {{x^2}\frac{d}{{dx}}y + y\frac{d}{{dx}}{x^2}} \right] + \left[ {x\frac{d}{{dx}}{y^2} + {y^2}\frac{d}{{dx}}x} \right] + 3{y^2}\frac{{dy}}{{dx}} = 0\] \[3{x^2} + {x^2}\frac{{dy}}{{dx}} + y\left( {2x} \right) + x\left( {2y} \right)\frac{{dy}}{{dx}} + {y^2} + 3{y^2}\frac{{dy}}{{dx}} = 0\] \[\frac{{dy}}{{dx}}\left( {{x^2} + 2xy + 3{y^2}} \right) = - \left( {3{x^2} + 2xy + {y^2}} \right)\] \[\frac{{dy}}{{dx}} = \frac{{ - \left( {3{x^2} + 2xy + {y^2}} \right)}}{{\left( {{x^2} + 2xy + 3{y^2}} \right)}}\]

Question (7)

sin2y + cosxy = π

Solution

\[\frac{d}{{dx}}{\sin ^2}x = 2\sin x\cos x\] \[\frac{d}{{dx}}\cos x = - \sin x\] \[\frac{d}{{dx}}uv = u\frac{{dv}}{{dx}} + v\frac{{du}}{{dx}}\] \[2\sin x\cos x = \sin 2x\] \[\frac{d}{{dx}}\left( c \right) = 0\] c= constant
sin2y + cosxy = π
Diff.w.r.t.x, \[2\sin y\cos y\frac{{dy}}{{dx}} + \left[ { - \sin \left( {xy} \right)} \right]\left[ {\frac{d}{{dx}}\left( {xy} \right)} \right] = 0\] \[\sin \left( {2y} \right)\frac{{dy}}{{dx}} - \sin xy\left[ {x\frac{{dy}}{{dx}} + y\frac{d}{{dx}}x} \right] = 0\] \[\sin \left( {2y} \right)\frac{{dy}}{{dx}} - x\sin xy\frac{{dy}}{{dx}} - y\sin xy = 0\] \[\frac{{dy}}{{dx}}\left( {\sin 2y - x\sin xy} \right) = y\sin xy\] \[\frac{{dy}}{{dx}} = \frac{{y\sin xy}}{{\sin 2y - x\sin xy}}\]

Question (8)

sin2x + cos2y = 1

Solution

\[\frac{d}{{dx}}{\sin ^2}x = 2\sin x\cos x\] \[\frac{d}{{dy}}{\cos ^2}x = - 2\sin x\cos x\] \[2\sin x\cos x = \sin 2x\]
sin2x + cos2y = 1
Diff.w.r.t.x, \[2\sin x\cos x + 2\cos y\left( { - \sin y} \right)\frac{{dy}}{{dx}} = 0\] \[\sin 2x - \sin 2y\frac{{dy}}{{dx}} = 0\] \[\sin 2y\frac{{dy}}{{dx}} = \sin 2x\] \[\frac{{dy}}{{dx}} = \frac{{\sin 2x}}{{\sin 2y}}\]

Question (9)

\[y = {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\]

Solution

\[\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = \sin 2\theta \] \[\frac{d}{{dx}}{\tan ^2}x = \frac{1}{{1 + {x^2}}}\]
let x=tanθ ⇒ θ = tan-1x \[y = {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\] \[y = {\sin ^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)\] \[y = {\sin ^{ - 1}}\left( {\sin 2\theta } \right)\] \[y = 2\theta \] \[y = {\tan ^{ - 1}}x\] differentiate w.r.t x \[\frac{{dy}}{{dx}} = \frac{2}{{1 + {x^2}}}\]

Question (10)

\[y = {\tan ^{ - 1}}\left( {\frac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right)\] \[ - \frac{1}{{\sqrt 3 }} < x < \frac{1}{{\sqrt 3 }}\]

Solution

\[\frac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }} = \tan 3\theta \] \[{\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta ,if\theta \in \left( {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right)\] \[\frac{d}{{dx}}{\tan ^{ - 1}}x = \frac{1}{{1 + {x^2}}}\]
Let x = tanθ ⇒ θ = tan-1x
\[\frac{{ - 1}}{{\sqrt 3 }} < x < \frac{1}{{\sqrt 3 }}\] \[\frac{{ - 1}}{{\sqrt 3 }} < \tan \theta < \frac{1}{{\sqrt 3 }}\] \[ - \tan \frac{\pi }{6} < \tan \theta < \tan \frac{\pi }{6}\] \[\frac{{ - \pi }}{6} < \theta < \frac{\pi }{6}\] \[\frac{{ - \pi }}{2} < 3\theta < \frac{\pi }{2}\] \[y = {\tan ^{ - 1}}\left( {\frac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right)\] \[y = {\tan ^{ - 1}}\left( {\tan 3\theta } \right)\quad ,3\theta \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\] \[y = 3\theta \] \[y = 3{\tan ^{ - 1}}x\] differentiate w.r.t x \[\frac{{dy}}{{dx}} = \frac{3}{{1 + {x^2}}}\]

Question (11)

\[y = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right),0 < x < 1\]

Solution

\[\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \cos 2\theta \] \[{\cos ^{ - 1}}\left( {\cos 2\theta } \right) = 2\theta \] if 2θ ∈ (0, π) \[\frac{d}{{dx}}{\tan ^{ - 1}}x = \frac{1}{{1 + {x^2}}}\]
Let x = tanθ ⇒ θ = tan-1x
0 < x < 1
0 < tanθ < 1
tan0 < tanθ < tan(π/4)
0 < θ < π/4
0 < 2θ < π/2     2θ ∈ (0, π/2)
\[y = {\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)\] \[y = {\cos ^{ - 1}}\left( {\cos 2\theta } \right)\] \[y = 2\theta \;\quad 2\theta \in \left( {0,\frac{\pi }{2}} \right)\] \[y = 2{\tan ^{ - 1}}x\;\] differentiate w.r.t x \[\frac{{dy}}{{dx}} = \frac{2}{{1 + {x^2}}}\]

Question (12)

\[y = {\sin ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right),0 < x < 1\]

Solution

\[\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \cos 2\theta \] \[\cos \theta = \sin \left( {\frac{\pi }{2} - \theta } \right)\] \[\frac{d}{{dx}}{\tan ^{ - 1}}x = \frac{1}{{1 + {x^2}}}\] \[{\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta ;\;\quad \theta \in \left( {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right)\] \[\frac{d}{{dx}}\left( c \right) = 0\]
Let x = tanθ ⇒ θ = tan-1x
0 < x < 1
0 < tanθ < 1
tan0 < tanθ < tan(π/4)
0 < θ < π/4
0 < 2θ < π/2
2θ ∈ (0, π/2)
2θ ∈ (0, π/2)
π/2 -2θ = (0, π/2) \[y = {\sin ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)\] \[y = {\sin ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)\] \[y = {\sin ^{ - 1}}\left( {\cos 2\theta } \right)\] \[y = {\sin ^{ - 1}}\left[ {\sin \left( {\frac{\pi }{2} - 2\theta } \right)} \right]\] \[y = \frac{\pi }{2} - 2\theta \] \[y = \frac{\pi }{2} - 2{\tan ^{ - 1}}x\] differentiate w.r.t x \[\frac{{dy}}{{dx}} = 0 - \frac{2}{{1 + {x^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{ - 2}}{{1 + {x^2}}}\]

Question (13)

\[y = {\cos ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right), - 1 < x < 1\]

Solution

\[\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = \sin 2\theta \] \[\sin \theta = \cos \left( {\frac{\pi }{2} - \theta } \right)\] cos-1(cosθ) = θ iif θ ∈ (0, π) \[\frac{d}{{dx}}{\tan ^{ - 1}}x = \frac{1}{{1 + {x^2}}}\]
Let x = tanθ ⇒ θ = tan-1x
-1 < x < 1
-1 < tanθ < 1
-tanπ/4 < tanθ <tanπ/4
-π/4 < θ <π/4
-π/2 < 2θ <π/2
∴ 2θ ∈ (-π/2 , π/2) \[y = {\cos ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\] \[y = {\cos ^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)\] \[y = {\cos ^{ - 1}}\left( {\sin 2\theta } \right)\] \[y = {\cos ^{ - 1}}\left( {\cos \left( {\frac{\pi }{2} - 2\theta } \right)} \right)\quad ,2\theta \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\] \[y = \frac{\pi }{2} - 2\theta \] \[y = \frac{\pi }{2} - 2{\tan ^{ - 1}}x\] differentiate w.r.t x \[\frac{{dy}}{{dx}} = \frac{{ - 2}}{{1 + {x^2}}}\]

Question (14)

\[y = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right), - \frac{1}{{\sqrt 2 }} < x < \frac{1}{{\sqrt 2 }}\]

Solution

2sinθcosθ = sin2θ
sin-1(sinθ) = θ if θ ∈ ( -π/2, π/2) \[\frac{d}{{dx}}{\sin ^{ - 1}}x = \frac{1}{{\sqrt {1 - {x^2}} }}\]
Let x = sinθ ⇒ = sin-1x
-1/√2 < x < 1/√2
-sin(π/4) < sinθ < sin(π/4)
-(π/4) < θ < (π/4)
-(π/2) < 2θ < (π/2)
2θ ∈ ( -π/2 , π/2) \[y = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)\] \[y = {\sin ^{ - 1}}\left( {2\sin \theta \sqrt {1 - {{\sin }^2}\theta } } \right)\] \[y = {\sin ^{ - 1}}\left( {2\sin \theta \cos \theta } \right)\] \[y = {\sin ^{ - 1}}\left( {\sin 2\theta } \right)\] \[y = 2\theta \] \[y = 2{\sin ^{ - 1}}x\] differentiate w.r.t x \[\frac{{dy}}{{dx}} = \frac{2}{{\sqrt {1 - {x^2}} }}\]

Question (15)

\[y = {\sec ^{ - 1}}\left( {\frac{1}{{2{x^2} - 1}}} \right),0 < x < \frac{1}{{\sqrt 2 }}\]

Solution

2cos2θ - 1 = cos2θ \[\frac{1}{{\cos \theta }} = \sec \theta \] sec-1(secθ) = θ if θ (0, π) \[\frac{d}{{dx}}{\cos ^{ - 1}}x = \frac{{ - 1}}{{\sqrt {1 - {x^2}} }}\]
\[0 < x < \frac{1}{{\sqrt 2 }}\] \[0 < \cos \theta < \frac{1}{{\sqrt 2 }} \Rightarrow \theta \in \left( {\frac{\pi }{4},\frac{\pi }{2}} \right)\] \[ \Rightarrow 2\theta \in \left( {\frac{\pi }{2},\pi } \right)\] \[y = {\sec ^{ - 1}}\left( {\frac{1}{{2{x^2} - 1}}} \right)\] \[y = {\sec ^{ - 1}}\left( {\frac{1}{{2{{\cos }^2}\theta - 1}}} \right)\] \[y = {\sec ^{ - 1}}\left( {\frac{1}{{\cos 2\theta }}} \right)\] \[y = {\sec ^{ - 1}}\left( {\sec 2\theta } \right)\] \[y = 2\theta \] \[y = 2{\cos ^{ - 1}}x\] differentiate w.r.t x \[\frac{{dy}}{{dx}} = \frac{{ - 2}}{{\sqrt {1 - {x^2}} }}\]
Exercise 5.2⇐
⇒Exercise 5.4