Chapter 13 of the NCERT Class 11 Mathematics textbook titled "Limits and Derivatives" introduces fundamental concepts in calculus focusing on the behavior of functions as they approach specific points and the concept of the derivatives. This chapter lays the groundwork for understanding the rate of the change of the functions and forms the basis for the more advanced calculus topics. The Miscellaneous Exercise in this chapter includes a variety of the problems designed to reinforce these concepts and provide practice in applying limits and derivatives.
Limits and Derivatives - Miscellaneous
The Miscellaneous Exercise in Chapter 13 covers diverse problems related to the limits and derivatives. It includes various types of questions such as the finding limits of functions evaluating the derivatives using the basic rules and applying derivative concepts to solve real-world problems. These exercises are intended to deepen students' understanding of how limits lead to derivatives and how to use these tools in different mathematical scenarios. The Mastery of these concepts is crucial for advancing in calculus and understanding its applications in various fields.
Question 1: Find the derivative of the following functions from the first principle:
(i) -x
Solution:
f(x) = -x
f(x+h) = -(x+h)
From the first principle,
f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} (\frac{-(x+h)-(-x)}{h})\\ f'(x) = \lim_{h \to 0} (\frac{-x-h+x}{h})\\ f'(x) = \lim_{h \to 0} (\frac{-h}{h})\\ f'(x) = \lim_{h \to 0} -1 f'(x) = -1
(ii) (-x)-1
Solution:
f(x) = (-x)-1 =
\frac{-1}{x} f(x+h) = (-(x+h))-1 =
\frac{-1}{x+h} From the first principle,
f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} (\frac{\frac{-1}{x+h}-(\frac{-1}{x})}{h})\\ f'(x) = \lim_{h \to 0} (\frac{\frac{-1}{x+h}+\frac{1}{x}}{h})\\ f'(x) = \lim_{h \to 0} (\frac{\frac{-x+(x+h)}{x(x+h)}}{h})\\ f'(x) = \lim_{h \to 0} (\frac{h}{hx(x+h)})\\ f'(x) = \lim_{h \to 0} (\frac{1}{x(x+h)})\\ f'(x) = (\frac{1}{x(x+0)})\\ f'(x) = (\frac{1}{x^2})
(iii) sin(x+1)
Solution:
f(x) = sin(x+1)
f(x+h) = sin((x+h)+1)
From the first principle,
f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} (\frac{sin((x+h)+1)-(sin(x+1))}{h}) Using the trigonometric identity,
sin A - sin B = 2 cos
(\frac{A+B}{2}) sin(\frac{A-B}{2})
f'(x) = \lim_{h \to 0} (\frac{2 cos (\frac{x+h+1+(x+1)}{2}) sin (\frac{x+h+1-(x+1)}{2}}{h})\\ f'(x) = \lim_{h \to 0} (\frac{2 cos (\frac{2x+h+2}{2}) sin (\frac{h}{2})}{h})\\ f'(x) = \lim_{h \to 0} (2 cos (\frac{2x+h+2}{2})) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{h})) Multiply and divide by 2, we have
f'(x) = 2 cos (\frac{2x+0+2}{2})) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{h})) \times \frac{2}{2}\\ f'(x) = 2 cos (x+1) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{\frac{h}{2}})) \times \frac{1}{2} f'(x) = cos (x+1) (1)
f'(x) = cos (x+1)
(iv) cos(x-\frac{\pi}{8})
Solution:
Here,
f(x) = cos(x-\frac{\pi}{8})\\ f(x+h) = cos((x+h)-\frac{\pi}{8}) From the first principle,
f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} \frac{cos((x+h)-\frac{\pi}{8})-cos(x-\frac{\pi}{8})}{h} Using the trigonometric identity,
cos a - cos b = -2 sin
(\frac{a+b}{2}) sin(\frac{a-b}{2})
f'(x) = \lim_{h \to 0} \frac{-2 sin (\frac{(x+h)-\frac{\pi}{8}+x-\frac{\pi}{8}}{2}) sin (\frac{(x+h)-\frac{\pi}{8}-(x-\frac{\pi}{8})}{2})}{h}\\ f'(x) = \lim_{h \to 0} \frac{-2 sin (\frac{(2x+h)-\frac{2\pi}{8}}{2}) sin (\frac{(x+h)-\frac{\pi}{8}-x+\frac{\pi}{8})}{2})}{h}\\ f'(x) = \lim_{h \to 0} \frac{-2 sin (\frac{2x+h-\frac{\pi}{4}}{2}) sin (\frac{h}{2})}{h} Multiplying and dividing by 2,
f'(x) = \lim_{h \to 0} \frac{-2 sin (\frac{2x+h-\frac{\pi}{4}}{2}) sin (\frac{h}{2})}{h} \times \frac{2}{2}\\ f'(x) = \lim_{h \to 0} (-2 sin (\frac{2x+h-\frac{\pi}{4}}{2})) \lim_{h \to 0}\frac{sin (\frac{h}{2})}{\frac{h}{2}} \times \frac{1}{2}\\ f'(x) = (-sin (\frac{2x+0-\frac{\pi}{4}}{2})) \lim_{h \to 0}\frac{sin (\frac{h}{2})}{\frac{h}{2}}\\ f'(x) = (-sin (x-\frac{\pi}{8})) (1)\\ f'(x) = -sin (x-\frac{\pi}{8})
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: (x+a)
Solution:
f(x) = x+a
Taking derivative both sides,
\frac{d}{dx}(f(x)) = \frac{d}{dx}(x+a)\\ f'(x) = \frac{d}{dx}(x)+\frac{d}{dx}(a) As, the derivative of xn is nxn-1 and derivative of constant is 0.
f'(x) = (1.x^{1-1})+0\\ f'(x) = (1.x^0)+0\\ f'(x) = 1
Question 3: (px+q) (\frac{r}{s}+s)
Solution:
f(x) = (px+q) (\frac{r}{s}+s) Taking derivative both sides,
\frac{d}{dx}(f(x)) = \frac{d}{dx}((px+q) (\frac{r}{s}+s)) Using the product rule, we have
(uv)' = uv'+u'v
f'(x) = (px+q) \frac{d}{dx}(\frac{r}{s}+s) + (\frac{r}{s}+s) \frac{d}{dx}(px+q) As, the derivative of xn is nxn-1 and derivative of constant is 0.
f'(x) = (px+q) (0) + (\frac{r}{s}+s) (\frac{d}{dx}(px)+\frac{d}{dx}(q))\\ f'(x) = 0 + (\frac{r}{s}+s) (p+0)\\ f'(x) = p(\frac{r}{s}+s)
Question 4: (ax+b) (cx+d)2
Solution:
f(x) = (ax+b) (cx+d)2
Taking derivative both sides,
\frac{d}{dx}(f(x)) = \frac{d}{dx}((ax+b) (cx+d)^2) Using the product rule, we have
(uv)' = uv'+u'v
f'(x) = (ax+b) \frac{d}{dx}(cx+d)^2 + (cx+d)^2 \frac{d}{dx}(ax+b) As, the derivative of xn is nxn-1 and derivative of constant is 0.
f'(x) = (ax+b) \frac{d}{dx}(c^2x^2+2(cx)(d)+d^2) + (cx+d)^2 (\frac{d}{dx}(ax)+\frac{d}{dx}(b))\\ f'(x) = (ax+b) (c^2(2x)+2cd+0) + (cx+d)^2 (a+0)\\ f'(x) = (ax+b) (2xc^2+2cd) + a(cx+d)^2
Question 5: \frac{ax+b}{cx+d}
Solution:
f(x) = \frac{ax+b}{cx+d} Taking derivative both sides,
\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{ax+b}{cx+d}) Using the quotient rule, we have
(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(cx+d) \frac{d}{dx}(ax+b) - (ax+b)\frac{d}{dx}(cx+d)}{(cx+d)^2} As, the derivative of xn is nxn-1 and derivative of constant is 0.
f'(x) = \frac{(cx+d) (a) - (ax+b)(c)}{(cx+d)^2}\\ f'(x) = \frac{acx+ad - acx-bc}{(cx+d)^2}\\ f'(x) = \frac{ad-bc}{(cx+d)^2}
Question 6: \frac{1+\frac{1}{x}}{1-\frac{1}{x}}
Solution:
f(x) = \frac{1+\frac{1}{x}}{1-\frac{1}{x}}\\ f(x) = \frac{\frac{x+1}{x}}{\frac{x-1}{x}}\\ f(x) = \frac{x+1}{x-1} Taking derivative both sides,
\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{x+1}{x-1}) Using the quotient rule, we have
(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(x-1) \frac{d}{dx}(x+1) - (x+1)\frac{d}{dx}(x-1)}{(x-1)^2} As, the derivative of xn is nxn-1 and derivative of constant is 0.
f'(x) = \frac{(x-1) (1) - (x+1)(1)}{(x-1)^2}\\ f'(x) = \frac{(x-1 - x-1)}{(x-1)^2}\\ f'(x) = \frac{-2}{(x-1)^2}\\
Question 7: \frac{1}{ax^2+bx+c}
Solution:
f(x) = \frac{1}{ax^2+bx+c} Taking derivative both sides,
\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{1}{ax^2+bx+c}) Using the quotient rule, we have
(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(ax^2+bx+c) \frac{d}{dx}(1) - (1)\frac{d}{dx}(ax^2+bx+c)}{(ax^2+bx+c)^2} As, the derivative of xn is nxn-1 and derivative of constant is 0.
f'(x) = \frac{(ax^2+bx+c) (0) - (1)(a(2x)+b+0)}{(ax^2+bx+c)^2}\\ f'(x) = \frac{0 - (2ax+b+0)}{(ax^2+bx+c)^2}\\ f'(x) = \frac{- (2ax+b+0)}{(ax^2+bx+c)^2}
Question 8: \frac{ax+b}{px^2+qx+r}
Solution:
f(x) = \frac{ax+b}{px^2+qx+r} Taking derivative both sides,
\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{ax+b}{px^2+qx+r}) Using the quotient rule, we have
(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(px^2+qx+r) \frac{d}{dx}(ax+b) - (ax+b)\frac{d}{dx}(px^2+qx+r)}{(px^2+qx+r)^2} As, the derivative of xn is nxn-1 and derivative of constant is 0.
f'(x) = \frac{(px^2+qx+r) (a+0) - (ax+b)\frac{d}{dx}(p(2x)+q+0)}{(px^2+qx+r)^2}\\ f'(x) = \frac{a(px^2+qx+r) - (ax+b)(2px+q)}{(px^2+qx+r)^2}\\ f'(x) = \frac{apx^2+qax+ra - (2apx^2+qax+2bpx+bq)}{(px^2+qx+r)^2}\\ f'(x) = \frac{apx^2+qax+ra - 2apx^2-qax-2bpx-bq)}{(px^2+qx+r)^2}\\ f'(x) = \frac{ra - apx^2-2bpx-bq)}{(px^2+qx+r)^2}
Question 9: \frac{px^2+qx+r}{ax+b}
Solution:
f(x) = \frac{px^2+qx+r}{ax+b} Taking derivative both sides,
\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{px^2+qx+r}{ax+b}) Using the quotient rule, we have
(\frac{u}{v})' = \frac{uv'-u'v}{u^2}\\ f'(x) = \frac{(ax+b) \frac{d}{dx}(px^2+qx+r) - (px^2+qx+r)\frac{d}{dx}(ax+b)}{(ax+b)^2} As, the derivative of xn is nxn-1 and derivative of constant is 0.
f'(x) = \frac{(ax+b) (p(2x)+q+0) - (px^2+qx+r)(a+0)}{(ax+b)^2}\\ f'(x) = \frac{(ax+b) (2px+q) - (px^2+qx+r)(a)}{(ax+b)^2}\\ f'(x) = \frac{2apx^2+qax+2bpx+bq - (apx^2+qax+ra}{(ax+b)^2}\\ f'(x) = \frac{2apx^2+qax+2bpx+bq - apx^2-qax-ra}{(ax+b)^2}\\ f'(x) = \frac{2apx^2+qax+2bpx+bq - apx^2-qax-ra}{(ax+b)^2}\\ f'(x) = \frac{apx^2 - ra +2bpx+bq)}{(ax+b)^2}
Question 10: \frac{a}{x^4}-\frac{b}{x^2}+cos x
Solution:
f(x) = \frac{a}{x^4}-\frac{b}{x^2}+cos x Taking derivative both sides,
\frac{d}{dx}(f(x)) = \frac{d}{dx}(\frac{a}{x^4}-\frac{b}{x^2}+cos x)\\ f'(x) = \frac{d}{dx}(\frac{a}{x^4})-\frac{d}{dx}(\frac{b}{x^2})+\frac{d}{dx}(cos x) As, the derivative of xn is nxn-1 and derivative of constant is 0.
f'(x) = a \frac{d}{dx}(x^{-4}) - b\frac{d}{dx}(x^{-2})+(- sin x)\\ f'(x) = a (-4x^{-4-1}) - b((-2)x^{-2-1})+(- sin x)\\ f'(x) = a (-4x^{-5}) - b((-2)x^{-3})+(- sin x)\\ f'(x) = -[4ax^{-5} - 2bx^{-3} + sin x]
Question 11: 4\sqrt{x} - 2
Solution:
f(x) = 4\sqrt{x} - 2 Taking derivative both sides,
\frac{d}{dx}(f(x)) = \frac{d}{dx}(4\sqrt{x} - 2)\\ f'(x) = \frac{d}{dx}(4\sqrt{x})-\frac{d}{dx}(2) As, the derivative of xn is nxn-1 and derivative of constant is 0.
f'(x) = 4 \frac{d}{dx}(x^{\frac{1}{2}})-0\\ f'(x) = 4 (\frac{1}{2}x^{\frac{1}{2}-1})\\ f'(x) = 2 (x^{-\frac{1}{2}})\\ f'(x) = \frac{2}{\sqrt{x}}
Question 12: (ax+b)n
Solution:
f(x) = (ax+b)n
f(x+h) = (a(x+h)+b)n
f(x+h) = (ax+ah+b)n
From the first principle,
f'(x) = \lim_{h \to 0} \frac{f(x+h)-(ax+b)^n}{h}\\ f'(x) = \lim_{h \to 0} \frac{(ax+ah+b)^n-(ax+b)^n}{h}\\ f'(x) = \lim_{h \to 0} \frac{(ax+b)^n (1+\frac{ah}{ax+b})^n-(ax+b)^n}{h}\\ f'(x) = (ax+b)^n \lim_{h \to 0} \frac{(1+\frac{ah}{ax+b})^n-1}{h} Using the binomial expansion, we have
f'(x) = (ax+b)^n \lim_{h \to 0} \frac{[1+n(\frac{ah}{ax+b}+\frac{n(n-1)}{2!}(\frac{ah}{(ax+b)^2}+.....]-1}{h}\\ f'(x) = (ax+b)^n \lim_{h \to 0} \frac{[n(\frac{ah}{ax+b}+\frac{n(n-1)a^2h^2}{2!(ax+b)^2}+.................]}{h}\\ f'(x) = (ax+b)^n \lim_{h \to 0} [n(\frac{a}{ax+b}+\frac{n(n-1)a^2h}{2!(ax+b)^2}+.................]\\ f'(x) = (ax+b)^n [n(\frac{a}{ax+b}+\frac{n(n-1)a^2(0)}{2!(ax+b)^2}+0+0+0......]\\ f'(x) = (ax+b)^n (\frac{na}{ax+b})\\ f'(x) = na (ax+b)^{n-1}
Question 13: (ax+b)n (cx+d)m
Solution:
f(x) = (ax+b)n (cx+d)m
Taking derivative both sides,
\frac{d}{dx}(f(x)) = \frac{d}{dx}((ax+b)^n (cx+d)^m) Using the product rule, we have
(uv)' = uv'+u'v
f'(x) = (ax+b)^n \frac{d}{dx}(cx+d)^m + (cx+d)^m \frac{d}{dx}(ax+b)^n Let's take, g(x) = (cx+d)m
g'(x) = \frac{d}{dx}(cx+d)^m g(x+h) = (c(x+h)+d)m
g(x+h) = (cx+ch+d)m
From the first principle,
g'(x) = \lim_{h \to 0} \frac{g(x+h)-g(x)}{h}\\ g'(x) = \lim_{h \to 0} \frac{(cx+ch+d)^m-(cx+d)^m}{h}\\ g'(x) = \lim_{h \to 0} \frac{(cx+d)^m (1+\frac{ch}{cx+d})^m-(cx+d)^m}{h}\\ g'(x) = (cx+d)^m \lim_{h \to 0} \frac{(1+\frac{ch}{cx+d})^m-1}{h} Using the binomial expansion, we have
g'(x) = (cx+d)^m \lim_{h \to 0} \frac{[1+m(\frac{ch}{cx+d}+\frac{m(m-1)}{2!}(\frac{ch}{(cx+d})^2+.....)]-1}{h}\\ g'(x) = (cx+d)^m \lim_{h \to 0} \frac{[m(\frac{ch}{cx+d}+\frac{m(nm-1)c^2h^2}{2!(cx+d)^2}+.................)]}{h}\\ g'(x) = (cx+d)^m \lim_{h \to 0} [m(\frac{c}{cx+d}+\frac{m(m-1)c^2h}{2!(cx+d)^2}+.................)]\\ g'(x) = (cx+d)^m [m(\frac{c}{cx+d}+\frac{m(m-1)c^2(0)}{2!(cx+d)^2}+0+0+0......)]\\ g'(x) = (cx+d)^m (\frac{mc}{cx+d})\\ g'(x) = mc (cx+d)^{m-1} So, as
f'(x) = (ax+b)^n \frac{d}{dx}(cx+d)^m + (cx+d)^m \frac{d}{dx}(ax+b)^n\\ f'(x) = (ax+b)^n (mc (cx+d)^{m-1}) + (cx+d)^m (na (ax+b)^{n-1})
Question 14: sin (x + a)
Solution:
f(x) = sin(x+a)
f(x+h) = sin((x+h)+a)
From the first principle,
f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ f'(x) = \lim_{h \to 0} (\frac{sin((x+h)+a)-(sin(x+a))}{h}) Using the trigonometric identity,
sin A - sin B = 2 cos
(\frac{A+B}{2}) sin(\frac{A-B}{2})
f'(x) = \lim_{h \to 0} (\frac{2 cos (\frac{x+h+a+(x+a)}{2}) sin (\frac{x+h+a-(x+a)}{2}}{h})\\ f'(x) = \lim_{h \to 0} (\frac{2 cos (\frac{2x+h+2a}{2}) sin (\frac{h}{2})}{h})\\ f'(x) = \lim_{h \to 0} (2 cos (\frac{2x+h+2a}{2})) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{h})) Multiply and divide by 2, we have
f'(x) = 2 cos (\frac{2x+0+2a}{2})) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{h})) \times \frac{2}{2}\\ f'(x) = 2 cos (x+a) \lim_{h \to 0} \frac{(sin (\frac{h}{2})}{\frac{h}{2}})) \times \frac{1}{2}\\ f'(x) = cos (x+a) (1)\\ f'(x) = cos (x+a)
Question 15: cosec x cot x
Solution:
f(x) = cosec x cot x
Taking derivative both sides,
\frac{d}{dx}(f(x)) = \frac{d}{dx}(cosec x \hspace{0.1cm}cot x) Using the product rule, we have
(uv)' = uv'+u'v
f'(x) = cot x \frac{d}{dx}(cosec x) + (cosec x) \frac{d}{dx}(cot x) f'(x) = cot x (-cot x cosec x) + (cosec x) (-cosec2 x)
f'(x) = - cot2 x cosec x - cosec3 x