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 $\frac{{dy}}{{dx}} = \frac{{ - 2}}{{\sqrt {1 - {x^2}} }}$