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

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
$\frac{{dy}}{{dx}} = \cos ecx\frac{d}{{dx}}\cot x + \cot x\frac{d}{{dx}}\cos ecx$ $= \cos ecx\left( { - \cos e{c^2}x} \right) + \cot x\left( { - \cos ecx\cot x} \right)$ $= - \cos e{c^3}x - \cos ecx{\cot ^2}x$