11th NCERT/CBSE Limits and Derivatives Exercise Miscellaneous Q1 to Q15

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Question (1)

Find the derivative of the following functions from first principle: (i) –x (ii) (–x)^–1
(iii) sin (x + 1) $\left( {iv} \right)\quad \cos \left( {x - \frac{\pi }{8}} \right)$

Solution

(i) \[f(x) = - x\] By the first principle, \[f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{ - \left( {x + h} \right) - \left( { - x} \right)}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{ - x - h + x}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{h}\] \[ = - 1\] (ii) \[f(x) = - {x^{ - 1}}\] By the first principle, \[f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{ - {{\left( {x + h} \right)}^{ - 1}} - {{\left( { - x} \right)}^{ - 1}}}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{ - \frac{1}{{x + h}} + \frac{1}{x}}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - x + x + h}}{{x\left( {x + h} \right)}}} \right]\] \[\mathop { = \lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{h}{{x\left( {x + h} \right)}}} \right]\] \[ = \frac{1}{{{x^2}}}\] (iii) \[f(x) = \sin \left( {x + 1} \right)\] By the first principle, \[f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{Sin\left( {x + h + 1} \right) - Sin\left( {x + 1} \right)}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{2\cos \left( {\frac{{x + h + 1 + x + 1}}{2}} \right)\sin \left( {\frac{{x + h + 1 - x - 1}}{2}} \right)}}{h}\] \[ = 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\cos \left( {\frac{{2x + h + 2}}{2}} \right)Sin\left( {\frac{h}{2}} \right)\] \[ = 2\left[ {\mathop {\lim }\limits_{h \to 0} \cos \left( {\frac{{2x + h + 2}}{2}} \right)} \right]\left[ {\mathop {\lim }\limits_{h \to 0} \frac{{Sin\left( {\frac{h}{2}} \right)}}{{\frac{h}{2}}}} \right] \times \frac{1}{2}\] \[ = 2\cos \left( {x + 1} \right) \times 1 \times \frac{1}{2}\] \[ = \cos \left( {x + 1} \right)\] (iv) \[f(x) = \cos \left( {x - \frac{\pi }{8}} \right)\] By the first principle, \[f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{\cos \left( {x + h - \frac{\pi }{8}} \right) - \cos \left( {x - \frac{\pi }{8}} \right)}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2\sin \left( {\frac{{x + h - \frac{\pi }{8} + x - \frac{\pi }{8}}}{2}} \right)\sin \left( {\frac{{x + h - \frac{\pi }{8} - x + \frac{\pi }{8}}}{2}} \right)}}{h}\] \[ = - 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\sin \left( {\frac{{2x + h - \frac{{2\pi }}{8}}}{2}} \right)Sin\left( {\frac{h}{2}} \right)\] \[ = - 2\left[ {\mathop {\lim }\limits_{h \to 0} \sin \left( {\frac{{2x + h - \frac{{2\pi }}{8}}}{2}} \right)} \right]\left[ {\mathop {\lim }\limits_{h \to 0} \frac{{Sin\left( {\frac{h}{2}} \right)}}{{\frac{h}{2}}}} \right] \times \frac{1}{2}\] \[ = - 2\sin \left( {x - \frac{\pi }{8}} \right) \times 1 \times \frac{1}{2}\] \[ = - \sin \left( {x - \frac{\pi }{8}} \right)\]

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

Question (2)

y = (x + a)

Solution

y = (x + a)
Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = \frac{d}{{dx}}(x + a)\] \[ = 1\]

Question (3)

\[\left( {px + q} \right)\left( {\frac{r}{x} + s} \right)\]

Solution

\[y = \left( {px + q} \right)\left( {\frac{r}{x} + s} \right)\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = \left( {px + q} \right)\frac{d}{{dx}}\left( {\frac{r}{x} + s} \right) + \left( {\frac{r}{x} + s} \right)\frac{d}{{dx}}\left( {px + q} \right)\] \[ = \left( {px + q} \right)\left( {\frac{{ - r}}{{{x^2}}}} \right) + \left( {\frac{r}{x} + s} \right)p\] \[ = \frac{{ - pr}}{x} - \frac{{qr}}{{{x^2}}} + \frac{{pr}}{x} + ps\] \[ = ps - \frac{{qr}}{{{x^2}}}\]

Question (4)

(ax + b) (cx + d)²

Solution

\[y = \left( {ax + b} \right){\left( {cx + d} \right)^2}\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = \left( {ax + b} \right)\frac{d}{{dx}}{\left( {cx + d} \right)^2} + {\left( {cx + d} \right)^2}\frac{d}{{dx}}\left( {ax + b} \right)\] \[ = \left( {ax + b} \right)2\left( {cx + d} \right)c + {\left( {cx + d} \right)^2}a\] \[ = \left( {cx + d} \right)\left[ {2c\left( {ax + b} \right) + a\left( {cx + d} \right)} \right]\] \[ = \left( {cx + d} \right)\left[ {2acx + 2bc + acx + ad} \right]\] \[ = \left( {cx + d} \right)\left[ {3acx + 2bc + ad} \right]\]

Question (5)

\[\frac{{ax + b}}{{cx + d}}\]

Solution

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

Question (6)

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

Solution

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

Question (7)

\[\frac{1}{{a{x^2} + bx + c}}\]

Solution

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

Question (8)

\[\frac{{ax + b}}{{p{x^2} + qx + r}}\]

Solution

\[y = \frac{{ax + b}}{{p{x^2} + qx + r}}\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = \frac{{\left( {p{x^2} + qx + r} \right)\frac{d}{{dx}}\left( {ax + b} \right) - \left( {ax + b} \right)\frac{d}{{dx}}\left( {p{x^2} + qx + r} \right)}}{{{{\left( {p{x^2} + qx + r} \right)}^2}}}\] \[ = \frac{{\left( {p{x^2} + qx + r} \right)a - \left( {ax + b} \right)\left( {2px + q} \right)}}{{{{\left( {p{x^2} + qx + r} \right)}^2}}}\] \[ = \frac{{ap{x^2} + aqx + ar - 2ap{x^2} - aqx - 2bpx - bq}}{{{{\left( {p{x^2} + qx + r} \right)}^2}}}\] \[ = \frac{{ - ap{x^2} - 2bpx + ar - bq}}{{{{\left( {p{x^2} + qx + r} \right)}^2}}}\]

Question (9)

\[\frac{{p{x^2} + qx + r}}{{ax + b}}\]

Solution

\[y = \frac{{p{x^2} + qx + r}}{{ax + b}}\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = \frac{{\left( {ax + b} \right)\frac{d}{{dx}}\left( {p{x^2} + qx + r} \right) - \left( {p{x^2} + qx + r} \right)\frac{d}{{dx}}\left( {ax + b} \right)}}{{{{\left( {ax + b} \right)}^2}}}\] \[ = \frac{{\left( {ax + b} \right)\left( {2px + q} \right) - \left( {p{x^2} + qx + r} \right)a}}{{{{\left( {ax + b} \right)}^2}}}\] \[ = \frac{{2ap{x^2} + aqx + 2bpx + bq - ap{x^2} - aqx - ar}}{{{{\left( {ax + b} \right)}^2}}}\] \[ = \frac{{ap{x^2} + 2bpx - ar + bq}}{{{{\left( {ax + b} \right)}^2}}}\]

Question (10)

\[\frac{a}{{{x^4}}} - \frac{b}{{{x^2}}} + \cos x\]

Solution

\[y = \frac{a}{{{x^4}}} - \frac{b}{{{x^2}}} + \cos x\] \[ = a{x^{ - 4}} + b{x^{ - 2}} + \cos x\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = a\frac{d}{{dx}}{x^{ - 4}} - b\frac{d}{{dx}}{x^{ - 2}} + \frac{d}{{dx}}\cos x\] \[ = a\left( { - 4{x^{ - 5}}} \right) - b\left( { - 2{x^{ - 3}}} \right) + \left( { - \sin x} \right)\] \[ = \frac{{ - 4a}}{{{x^5}}} + \frac{{2b}}{{{x^3}}} - \sin x\]

Question (11)

\[4\sqrt x - 2\]

Solution

\[y = 4\sqrt x - 2\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = 4\frac{d}{{dx}}{x^{1/2}} - \frac{d}{{dx}}2\] \[ = 4\left( {\frac{1}{2}{x^{ - 1/2}}} \right) - 0\] \[ = \frac{2}{{\sqrt x }}\]

Question (12)

(ax+b)ⁿ

Solution

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

Question (13)

(ax+b)ⁿ(cx+d)^m

Solution

\[y = {\left( {ax + b} \right)^n}{\left( {cx + d} \right)^m}\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = {\left( {ax + b} \right)^n}\frac{d}{{dx}}{\left( {cx + d} \right)^m} + {\left( {cx + d} \right)^m}\frac{d}{{dx}}{\left( {ax + b} \right)^n}\] \[ = {\left( {ax + b} \right)^n}m{\left( {cx + d} \right)^{m - 1}}\frac{d}{{dx}}\left( {cx + d} \right) + {\left( {cx + d} \right)^m}n{\left( {ax + b} \right)^{n - 1}}\frac{d}{{dx}}\left( {ax + b} \right)\] \[ = mc{\left( {ax + b} \right)^n}{\left( {cx + d} \right)^{m - 1}} + n{\left( {cx + d} \right)^m}{\left( {ax + b} \right)^{n - 1}}\left( a \right)\] \[ = {\left( {ax + b} \right)^{n - 1}}{\left( {cx + d} \right)^{m - 1}}\left[ {mc\left( {ax + b} \right) + na\left( {cx + d} \right)} \right]\] \[ = {\left( {ax + b} \right)^{n - 1}}{\left( {cx + d} \right)^{m - 1}}\left[ {amcx + bmc + nacx + nad} \right]\]

Question (14)

sin(x+a)

Solution

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

Question (15)

cosec x cot x

Solution

\[y = \cos ecx\cot x\] Diff.w.r.t.x, we get
\[\frac{{dy}}{{dx}} = \cos ecx\frac{d}{{dx}}\cot x + \cot x\frac{d}{{dx}}\cos ecx\] \[ = \cos ecx\left( { - \cos e{c^2}x} \right) + \cot x\left( { - \cos ecx\cot x} \right)\] \[ = - \cos e{c^3}x - \cos ecx{\cot ^2}x\]