11th NCERT/CBSE Limits and Derivatives Exercise Miscellaneous Q16 to Q30

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Question (16)

\[\frac{{\cos x}}{{1 + \sin x}}\]

Solution

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

Question (17)

\[\frac{{\sin x + \cos x}}{{\sin x - \cos x}}\]

Solution

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

Question (18)

\[\frac{{\sec x - 1}}{{\sec x + 1}}\]

Solution

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

Question (19)

sinⁿx

Solution

\[y = {\sin ^n}x\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\left( {\sin x} \right)^n}\] \[ = n{\left( {\sin x} \right)^{n - 1}}\frac{d}{{dx}}\left( {\sin x} \right)\] \[ = n{\left( {\sin x} \right)^{n - 1}}\cos x\]

Question (20)

\[\frac{{a + b\sin x}}{{c + d\cos x}}\]

Solution

\[y = \frac{{a + b\sin x}}{{c + d\cos x}}\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = \frac{{\left( {c + d\cos x} \right)\frac{d}{{dx}}\left( {a + b\sin x} \right) - \left( {a + b\sin x} \right)\frac{d}{{dx}}\left( {c + d\cos x} \right)}}{{{{\left( {c + d\cos x} \right)}^2}}}\] \[ = \frac{{\left( {c + d\cos x} \right)\left( {b\cos x} \right) - \left( {a + b\sin x} \right)\left( { - d\sin x} \right)}}{{{{\left( {c + d\cos x} \right)}^2}}}\] \[ = \frac{{bc\cos x + bd{{\cos }^2}x + ad\sin x + bd{{\sin }^2}x}}{{{{\left( {c + d\cos x} \right)}^2}}}\] \[ = \frac{{bc\cos x + ad\sin x + bd\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{{{\left( {c + d\cos x} \right)}^2}}}\] \[ = \frac{{bc\cos x + ad\sin x + bd}}{{{{\left( {c + d\cos x} \right)}^2}}}\]

Question (21)

\[\frac{{\sin \left( {x + a} \right)}}{{\cos x}}\]

Solution

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

Question (22)

x⁴ (5sinx - 3cosx)

Solution

\[y = {x^4}\left( {5\sin x - 3\cos x} \right)\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = {x^4}\frac{d}{{dx}}\left( {5\sin x - 3\cos x} \right) + \left( {5\sin x - 3\cos x} \right)\frac{d}{{dx}}{x^4}\] \[ = {x^4}\left( {5\cos x + 3\sin x} \right) + \left( {5\sin x - 3\cos x} \right)4{x^3}\] \[ = {x^3}\left[ {5x\cos x + 3x\sin x + 20\sin x - 12\cos x} \right]\]

Question (23)

(x² + 1) cosx

Solution

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

Question (24)

(ax² + sin x) (p + qcosx)

Solution

\[y = \left( {a{x^2} + \sin x} \right)\left( {p + q\cos x} \right)\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = \left( {a{x^2} + \sin x} \right)\frac{d}{{dx}}\left( {p + q\cos x} \right) + \left( {p + q\cos x} \right)\frac{d}{{dx}}\left( {a{x^2} + \sin x} \right)\] \[ = \left( {a{x^2} + \sin x} \right)\left( { - q\sin x} \right) + \left( {p + q\cos x} \right)\left( {2ax + \cos x} \right)\] \[ = - aq{x^2}\sin x - q{\sin ^2}x + 2apx + p\cos x + 2aqx\cos x + {\cos ^2}x\]

Question (25)

(x + cosx) (x - tanx)

Solution

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

Question (26)

\[\frac{{4x + 5\sin x}}{{3x + 7\cos x}}\]

Solution

\[y = \frac{{4x + 5\sin x}}{{3x + 7\cos x}}\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = \frac{{\left( {3x + 7\cos x} \right)\frac{d}{{dx}}\left( {4x + 5\sin x} \right) - \left( {4x + 5\sin x} \right)\frac{d}{{dx}}\left( {3x + 7\cos x} \right)}}{{{{\left( {3x + 7\cos x} \right)}^2}}}\] \[ = \frac{{\left( {3x + 7\cos x} \right)\left( {4 + 5\cos x} \right) - \left( {4x + 5\sin x} \right)\left( {3 - 7\sin x} \right)}}{{{{\left( {3x + 7\cos x} \right)}^2}}}\] \[ = \frac{{12x + 28\cos x + 15x\cos x + 35{{\cos }^2}x - 12x + 28x\sin x - 15\sin x + 35{{\sin }^2}x}}{{{{\left( {3x + 7\cos x} \right)}^2}}}\] \[ = \frac{{35\left( {{{\cos }^2}x + {{\sin }^2}x} \right) + 28\cos x + 15x\cos x + 28x\sin x - 15\sin x}}{{{{\left( {3x + 7\cos x} \right)}^2}}}\] \[\begin{array}{l} = \frac{{35 + 28\cos x + 15x\cos x + 28x\sin x - 15\sin x}}{{{{\left( {3x + 7\cos x} \right)}^2}}}\\\end{array}\]

Question (27)

\[\frac{{{x^2}\cos \left( {\frac{\pi }{4}} \right)}}{{\sin x}}\]

Solution

\[y = \frac{{{x^2}\cos \frac{\pi }{4}}}{{\sin x}}\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = \frac{{\left( {\sin x} \right)\frac{d}{{dx}}\left( {{x^2}\cos \frac{\pi }{4}} \right) - \left( {{x^2}\cos \frac{\pi }{4}} \right)\frac{d}{{dx}}\left( {\sin x} \right)}}{{{{\left( {\sin x} \right)}^2}}}\] \[ = \frac{{\left( {\sin x} \right)\left( {2x\cos \frac{\pi }{4}} \right) - \left( {{x^2}\cos \frac{\pi }{4}} \right)\left( {\cos x} \right)}}{{{{\left( {\sin x} \right)}^2}}}\] \[ = \frac{{x\cos \frac{\pi }{4}\left[ {2\sin x - x\cos x} \right]}}{{{{\sin }^2}x}}\]

Question (28)

\[\frac{x}{{1 + \tan x}}\]

Solution

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

Question (29)

(x + secx) (x - tanx)

Solution

\[y = \left( {x + \sec x} \right)\left( {x - \tan x} \right)\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = \left( {x + \sec x} \right)\frac{d}{{dx}}\left( {x - \tan x} \right) + \left( {x - \tan x} \right)\frac{d}{{dx}}\left( {x + secx} \right)\] \[ = \left( {x + \sec x} \right)\left( {1 - {{\sec }^2}x} \right) + \left( {x - \tan x} \right)\left( {1 + \sec x\tan x} \right)\]

Question (30)

\[\frac{x}{{{{\sin }^n}x}}\]

Solution

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