11th NCERT/CBSE Limits and Derivatives Exercise 12.2 Questions 11
Hi

Question (1)

Find the derivative of x2 – 2 at x = 10

Solution

Let f(x) = x2 - 2
Diff. w.r.t.x, we get,
f'( x) = 2x
at x = 10
f'( 10) = 2 (10)
= 20

Question (2)

Find the derivative of 99x at x = 100.

Solution

Let f(x) = 99x
Diff. w.r.t.x, we get,
f'( x) = 99
At x = 100
f'(100) = 99

Question (3)

Find the derivative of x at x = 1.

Solution

f(x) = x
Diff. w.r.t.x, we get,
f'( x) = 1
At x = 1
f'(1) = 1

Question (4)

Find the derivative of the following functions from first principle.
(i) x3 – 27 (ii) (x – 1) (x – 2)
\[\left( {iii} \right)\frac{1}{{{x^2}}}\quad \quad \left( {iv} \right)\frac{{x + 1}}{{x - 1}}\]

Solution

(i) x3 - 27
By first principle , derivative of f(x) is,
\[f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{\left( {x + h} \right)}^3} - 27} \right] - \left[ {{x^3} - 27} \right]}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^3} - {x^3}}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {x + h - x} \right)\left[ {{{\left( {x + h} \right)}^2} + x\left( {x + h} \right) + {x^2}} \right]}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{h\left[ {{{\left( {x + h} \right)}^2} + x\left( {x + h} \right) + {x^2}} \right]}}{h}\] \[ = {x^2} + {x^2} + {x^2}\] \[ = 3{x^2}\] (ii) f(x) = (x - 1 )(x - 2)
\[f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\left( {x + h - 1} \right)\left( {x + h - 2} \right)} \right] - \left( {x - 1} \right)\left( {x - 2} \right)}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} + xh - 2x + hx + {h^2} - 2h - x - h + 2 - {x^2} + 3x - 2}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{2xh + {h^2} - 3h}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{h\left[ {2x + h - 3} \right]}}{h}\] \[ = 2x + 0 - 3\] \[ = 2x - 3\] \[\left( {iii} \right)f(x) = \frac{1}{{{x^2}}}\] \[f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{{{\left( {x + h} \right)}^2}}} - \frac{1}{{{x^2}}}} \right]\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} - {{\left( {x + h} \right)}^2}}}{{h{x^2}{{\left( {x + h} \right)}^2}}}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} - {x^2} - 2xh - {h^2}}}{{h{x^2}{{\left( {x + h} \right)}^2}}}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{h\left[ { - 2x - h} \right]}}{{h{x^2}{{\left( {x + h} \right)}^2}}}\] \[ = \frac{{ - 2x}}{{{x^2}{{\left( {x + 0} \right)}^2}}}\] \[ = \frac{{ - 2x}}{{{x^4}}} = \frac{{ - 2}}{{{x^3}}}\] (iv) \[f(x) = \frac{{x + 1}}{{x - 1}}\quad \] \[f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{x + h + 1}}{{\left( {x + h - 1} \right)}} - \frac{{x + 1}}{{x - 1}}} \right]\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {x + h + 1} \right)\left( {x - 1} \right) - \left( {x + 1} \right)\left( {x + h - 1} \right)}}{{h\left( {x - 1} \right)\left( {x + h - 1} \right)}}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{x^2} - x + hx - h + x - 1} \right] - \left[ {{x^2} + xh - x + x + h - 1} \right]}}{{h\left( {x - 1} \right)\left( {x + h - 1} \right)}}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2h}}{{h\left( {x - 1} \right)\left( {x + h - 1} \right)}}\] \[ = \frac{{ - 2}}{{\left( {x - 1} \right)\left( {x + 0 - 1} \right)}}\] \[ = \frac{{ - 2}}{{{{\left( {x - 1} \right)}^2}}}\]

Question (5)

For the function
\[f\left( x \right) = \frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + .... + \frac{{{x^2}}}{2} + x + 1\] Prove that f'(1) = 100f'(x)

Solution

\[f\left( x \right) = \frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + .... + \frac{{{x^2}}}{2} + x + 1\] Diff.w.r.t.x, we get
\[f'\left( x \right) = \frac{{100{x^{99}}}}{{100}} + \frac{{99{x^{98}}}}{{99}} + ... + \frac{{2x}}{2} + 1 + 0\] \[ = {x^{99}} + {x^{98}} + ... + x + 1\] \[LHS = f'(1)\] \[ = {1^{99}} + {1^{98}} + ...1 + 1\] \[ = 100\] \[RHS = 100f'(0)\] \[ = 100\left[ {0 + 0 + ...0 + 1} \right]\] \[ = 100\] \[ = LHS\]

Question (6)

Find the derivative of xn + axn-1 + a2xn-1+ ...+ an-1x + an
for some fixed real number a.

Solution

\[f(x) = {x^n} + a{x^{n - 1}} + {a^2}{x^{n - 2}} + ......{a^n}\] Diff.w.r.t.x, we get,
\[f'\left( x \right) = n{x^{n - 1}} + a\left( {n - 1} \right){x^{n - 2}} + {a^2}\left( {n - 2} \right){x^{n - 3}} + .....{a^{n - 1}}\]

Question (7)

For some constants a and b, find the derivative of (i) (x – a) (x – b) (ii) (ax2 + b)2 $\left( {iii} \right)\frac{{x - a}}{{x - b}}$

Solution

(i) \[f(x) = \left( {x - a} \right)\left( {x - b} \right)\] Diff.w.r.t.x, we get
\[f'\left( x \right) = \left( {x - a} \right)\frac{d}{{dx}}\left( {x - b} \right) + \left( {a - b} \right)\frac{d}{{dx}}\left( {x - a} \right)\] \[ = \left( {x - a} \right)\left( 1 \right) + \left( {x - b} \right)\left( 1 \right)\] \[ = x - a + x - b\] \[ = 2x - a - b\] (ii) \[f(x) = {\left( {a{x^2} + b} \right)^2}\] Diff.w.r.t.x, we get
\[f'\left( x \right) = 2\left( {a{x^2} + b} \right)\frac{d}{{dx}}\left( {a{x^2} + b} \right)\] \[ = 2\left( {a{x^2} + b} \right)\left( {2ax + 0} \right)\] \[ = 4ax\left( {a{x^2} + b} \right)\] (iii) \right)\frac{{x - a}}{{x - b}} Diff.w.r.t.x, we get
\[f'\left( x \right) = \frac{{\left( {x - b} \right)\frac{d}{{dx}}\left( {x - a} \right) - \left( {x - a} \right)\frac{d}{{dx}}\left( {x - b} \right)}}{{{{\left( {x - b} \right)}^2}}}\] \[ = \frac{{\left( {x - b} \right)\left( {1 - 0} \right) - \left( {x - a} \right)\left( {1 - 0} \right)}}{{{{\left( {x - b} \right)}^2}}}\] \[ = \frac{{x - b - x + a}}{{{{\left( {x - b} \right)}^2}}}\] \[ = \frac{{a - b}}{{{{\left( {x - b} \right)}^2}}}\]

Question (8)

Find the derivative of $\frac{{{x^n} - {a^n}}}{{x - a}}$ for some constant a.

Solution

\[f(x) = \frac{{{x^n} - {a^n}}}{{x - a}}\] Diff.w.r.t.x, we get
\[f'\left( x \right) = \frac{{\left( {x - a} \right)\frac{d}{{dx}}\left( {{x^n} - {a^n}} \right) - \left( {{x^n} - {a^n}} \right)\frac{d}{{dx}}\left( {x - a} \right)}}{{{{\left( {x - a} \right)}^2}}}\] \[ = \frac{{\left( {x - a} \right)\left( {n{x^{n - 1}} - 0} \right) - \left( {{x^n} - {a^n}} \right)\left( {1 - 0} \right)}}{{{{\left( {x - a} \right)}^2}}}\] \[ = \frac{{n{x^n} - an{x^{n - 1}} - {x^n} + {a^n}}}{{{{\left( {x - a} \right)}^2}}}\]

Question (9)

Find the derivative of
$\text{(i)}\;\left( i \right)2x - \frac{3}{4}$
(ii) (5x3 + 3x – 1) (x – 1)
(iii) x–3 (5 + 3x)
(iv) x5 (3 – 6x–9)
(v) x–4 (3 – 4x–5)
$\left( {iv} \right)\quad \frac{2}{{x + 1}} - \frac{{{x^2}}}{{3x - 1}}$

Solution

\[\left( i \right)2x - \frac{3}{4}\] Diff.w.r.t.x, we get
\[f'\left( x \right) = 2 - 0\] \[ = 2\] (ii) \[f\left( x \right) = ({\rm{5}}{x^{\rm{3}}} + {\rm{ 3}}x--{\rm{ 1}}){\rm{ }}(x--{\rm{ 1}})\] Diff.w.r.t.x, we get
\[f'\left( x \right) = ({\rm{5}}{x^{\rm{3}}} + {\rm{ 3}}x--{\rm{ 1}})\frac{d}{{dx}}\left( {x - 1} \right) + \left( {x - 1} \right)\frac{d}{{dx}}({\rm{5}}{x^{\rm{3}}} + {\rm{ 3}}x--{\rm{ 1}})\] \[ = ({\rm{5}}{x^{\rm{3}}} + {\rm{ 3}}x--{\rm{ 1}})\left( 1 \right) + \left( {x - 1} \right)\left( {15{x^2} + 3} \right)\] \[ = {\rm{5}}{x^{\rm{3}}} + {\rm{ 3}}x--{\rm{ 1 + 15}}{{\rm{x}}^3} + 3x - 15{x^2} - 3\] \[ = 20{x^{\rm{3}}} - 15{x^2} + {\rm{ 6}}x--{\rm{ 4}}\] \[(iii)f\left( x \right) = {x^{ - 3}}\left( {5 + 3x} \right)\] \[ = 5{x^{ - 3}} + 3{x^{ - 2}}\] Diff.w.r.t.x, we get
\[f'\left( x \right) = 5\frac{d}{{dx}}{x^{ - 3}} + 3\frac{d}{{dx}}{x^{ - 2}}\] \[ = 5\left( { - 3} \right){x^{ - 4}} + 3\left( { - 2} \right){x^{ - 3}}\] \[ = - 15{x^{ - 4}} - 6{x^{ - 3}}\] \[(iv)f\left( x \right) = {x^5}\left( {3 - 6{x^{ - 9}}} \right)\] \[ = 3{x^5} - 6{x^{ - 4}}\] Diff.w.r.t.x, we get
\[f'\left( x \right) = 3\frac{d}{{dx}}{x^5} - 6\frac{d}{{dx}}{x^{ - 4}}\] \[f'\left( x \right) = 3\left( {5{x^4}} \right) - 6\left( { - 4{x^{ - 5}}} \right)\] \[ = 15{x^4} + 24{x^{ - 5}}\] \[(v)f\left( x \right) = {x^{ - 4}}\left( {3 - 4{x^{ - 5}}} \right)\] \[ = 3{x^{ - 4}} - 4{x^{ - 9}}\] Diff.w.r.t.x, we get
\[f'\left( x \right) = 3\frac{d}{{dx}}{x^{ - 4}} - 4\frac{d}{{dx}}{x^{ - 9}}\] \[ = 3\left( { - 4} \right){x^{ - 5}} - 4\left( { - 9} \right){x^{ - 10}}\] \[ = - 12{x^{ - 5}} + 36{x^{ - 10}}\] \[(vi)f\left( x \right) = \frac{2}{{x + 1}} - \frac{{{x^2}}}{{3x - 1}}\] Diff.w.r.t.x, we get \[f'\left( x \right) = 2\frac{d}{{dx}}{\left( {x + 1} \right)^{ - 1}} - \frac{d}{{dx}}\frac{{{x^2}}}{{3x - 1}}\] \[f'\left( x \right) = 2\left( { - 1} \right){\left( {x + 1} \right)^{ - 2}} - \left[ {\frac{{\left( {3x - 1} \right)\frac{d}{{dx}}{x^2} - {x^2}\frac{d}{{dx}}\left( {3x - 1} \right)}}{{{{\left( {3x - 1} \right)}^2}}}} \right]\] \[ = \frac{{ - 2}}{{{{\left( {x + 1} \right)}^2}}} - \left[ {\frac{{\left( {3x - 1} \right)2x - {x^2}\left( 3 \right)}}{{{{\left( {3x - 1} \right)}^2}}}} \right]\] \[ = \frac{{ - 2}}{{{{\left( {x + 1} \right)}^2}}} - \left[ {\frac{{6{x^2} - 2x - 3{x^2}}}{{{{\left( {3x - 1} \right)}^2}}}} \right]\] \[ = \frac{{ - 2}}{{{{\left( {x + 1} \right)}^2}}} - \frac{{3{x^2} - 2x}}{{{{\left( {3x - 1} \right)}^2}}}\]

Question (10)

Find the derivative of cos x from first principle.

Solution

\[f\left( x \right) = \cos x\] By First principle, \[f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{\cos \left( {x + h} \right) - \cos x}}{h}\] \[ = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2\sin \left( {\frac{{x + h + x}}{2}} \right)\sin \left( {\frac{{x + h - x}}{2}} \right)}}{h}\] \[ = - 2\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{h}{2}} \right)}}{h}\] \[ = - 2\mathop {\lim }\limits_{h \to 0} \sin \left( {\frac{{2x + h}}{2}} \right)\left[ {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\frac{h}{2}}}} \right] \times \frac{1}{2}\] \[ = - 2\sin \left( {\frac{{2x + 0}}{2}} \right)\left( 1 \right) \times \frac{1}{2}\] \[ = - \sin x\]

Question (11)

Find the derivative of the following functions: (i) sin x cos x (ii) sec x (iii) 5 sec x + 4 cos x
(iv) cosec x (v) 3cot x + 5cosec x
(vi) 5sin x – 6cos x + 7 (vii) 2tan x – 7sec x

Solution

\[(i)f(x) = \sin x\cos x\] Diff.w.r.t.x, we get
\[f'(x) = \sin x\frac{d}{{dx}}\cos x + \cos x\frac{d}{{dx}}\sin x\] \[ = \sin x\left( { - \sin x} \right) + \cos x\left( {\cos x} \right)\] \[ = {\cos ^2}x - {\sin ^2}x\] \[ = \cos (2x)\] \[(ii)f(x) = \sec x\] Diff.w.r.t.x, we get
\[f'(x) = \sec x\tan x\] \[(iii)f(x) = 5\sec x + 4\cos x\] Diff.w.r.t.x, we get
\[f'\left( x \right) = 5\frac{d}{{dx}}\sec x + 4\frac{d}{{dx}}\cos x\] \[ = 5\sec x\tan x + 4\left( { - \sin x} \right)\] \[ = 5\sec x\tan x - 4\sin x\] \[(iv)f(x) = \cos ecx\] Diff.w.r.t.x, we get
\[f'(x) = - \cos ecx\cot x\] \[(v)f(x) = 3\cot x + 5\cos ecx\] Diff.w.r.t.x, we get
\[f'(x) = 3\frac{d}{{dx}}\cot x + 5\frac{d}{{dx}}\cos ecx\] \[ = 3\left( { - \cos e{c^2}x} \right) + 5\left( { - \cos ecx\cot x} \right)\] \[ = - 3\cos e{c^2}x - 5\cos ecx\cot x\] \[(vi)f(x) = 5\sin x - 6\cos x + 7\] Diff.w.r.t.x, we get
\[f'(x) = 5\frac{d}{{dx}}\sin x - 6\frac{d}{{dx}}\cos x + \frac{d}{{dx}}7\] \[ = 5\cos x - 6\left( { - \sin x} \right) + 0\] \[ = 5\cos x + 6\sin x\] \[(vii)f(x) = 2\tan x - 7\sec x\] Diff.w.r.t.x, we get
\[f'\left( x \right) = 2\frac{d}{{dx}}\tan x - 7\frac{d}{{dx}}\sec x\] \[ = 2{\sec ^2}x - 7\left( {\sec x\tan x} \right)\] \[ = 2{\sec ^2}x - 7\sec x\tan x\]
13.1(Q17 t0 Q32)⇐
Miscellaneous 13