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 + by
2 = cosy
Solution
\[\frac{d}{{dx}}{x^n} = n{x^{n - 1}}\]
\[\frac{d}{{dx}}\cos x = - \sin x\]
ax + by
2 = 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 + y
2 = 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 + y
2 = 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)
x
2+xy+y
2=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
x
2+xy+y
2=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)
x
3 + x
2y +xy
2 + y
3 = 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
x
3 + x
2y +xy
2 + y
3 = 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)
sin
2y + 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
sin
2y + 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)
sin
2x + cos
2y = 1
Solution
\[\frac{d}{{dx}}{\sin ^2}x = 2\sin x\cos x\]
\[\frac{d}{{dy}}{\cos ^2}x = - 2\sin x\cos x\]
\[2\sin x\cos x = \sin 2x\]
sin
2x + cos
2y = 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
\[\frac{{dy}}{{dx}} = \frac{{ - 2}}{{\sqrt {1 - {x^2}} }}\]