Question 12 Marks
Find the derivative of function $\frac{{p{x^2} + qx + r}}{{ax + b}}$ (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).
AnswerHere $f(x) = \frac{{p{x^2} + qx + r}}{{ax + b}}$
$\therefore f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{p{x^2} + qx + r}}{{ax + b}}} \right]$
$= \frac{{(ax + b)\frac{d}{{dx}}(p{x^2} + qx + r) - (p{x^2} + qx + r)\frac{d}{{dx}}(ax + b)}}{{{{(ax + b)}^2}}}$
$= \frac{{(ax + b)(2px + q) - (p{x^2} + qx + r)(a)}}{{{{(ax + b)}^2}}}$
$= \frac{{2ap{x^2} + aqx + 2bpx + bq + ap{x^2} - aqx - ar}}{{{{(ax + b)}^2}}}$
$ = \frac{{ap{x^2} + 2bpx + bq - ar}}{{{{(ax + b)}^2}}}$
View full question & answer→Question 22 Marks
Find the derivative of function $\frac{{ax + b}}{{p{x^2} + qx + r}}$ (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):.
Answer$f(x) = \frac{{ax + b}}{{p{x^2} + qx + r}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{ax + b}}{{p{x^2} + qx + r}}} \right]$
$= \frac{{(p{x^2} + qx + r)\frac{d}{{dx}}(ax + b) - (ax + b)\frac{d}{{dx}}(p{x^2} + qx + r)}}{{{{(p{x^2} + qx + r)}^2}}}$
$ = \frac{{(p{x^2} + qx + r)(a) - (ax + b)(2px + q)}}{{{{(p{x^2} + qx + r)}^2}}}$
$ = \frac{{ap{x^2} + aqx + ar - 2ap{x^2} - aqx - 2bpx - bq}}{{{{(p{x^2} + qx + r)}^2}}}$
$= \frac{{ - ap{x^2} - 2bpx + ar - bq}}{{{{(p{x^2} + qx + r)}^2}}}$
View full question & answer→Question 32 Marks
Find the derivative of function $\frac{1}{{a{x^2} + bx + c}}$(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).
AnswerHere $f(x) = \frac{1}{{a{x^2} + bx + c}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left( {\frac{1}{{a{x^2} + bx + c}}} \right)$
$ = \frac{{(a{x^2} + bx + c)\frac{d}{{dx}}(1) - 1.\frac{d}{{dx}}(a{x^2} + bx + c)}}{{{{(a{x^2} + bx + c)}^2}}}$
$ = \frac{{(a{x^2} + bx + c)(0) - 1(2ax + b)}}{{{{(a{x^2} + bx + c)}^2}}}$$ = \frac{{ - (2ax + b)}}{{{{(a{x^2} + bx + c)}^2}}}$
View full question & answer→Question 42 Marks
Find the derivative of function $\frac{{1 + \frac{1}{x}}}{{1 - \frac{1}{x}}}$ (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).
AnswerHere $f(x) = \frac{{1 + \frac{1}{x}}}{{1 - \frac{1}{x}}} = \frac{{x + 1}}{{x - 1}}$
$f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{x + 1}}{{x - 1}}} \right]$
$ = \frac{{(x - 1)\frac{d}{{dx}}(x + 1) - (x + 1)\frac{d}{{dx}}(x - 1)}}{{{{(x - 1)}^2}}}$
$= \frac{{(x - 1) \times 1 - (x + 1) \times 1}}{{{{(x - 1)}^2}}}$
$ = \frac{{x - 1 - x - 1}}{{{{(x - 1)}^2}}} = \frac{{ - 2}}{{{{(x - 1)}^2}}},\;x \ne 0,\;1.$
View full question & answer→Question 52 Marks
Find the derivative of function $\frac{{ax + b}}{{cx + d}}$ (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).
AnswerHere $f(x) = \frac{{ax + b}}{{cx + d}}$
$\therefore \;f'(x) = \frac{d}{{dx}}\left[ {\frac{{ax + b}}{{cx + d}}} \right]$
$= \frac{{(cx + d)\frac{d}{{dx}}(ax + b) - (ax + b)\frac{d}{{dx}}(cx + d)}}{{{{(cx + d)}^2}}}$
$= \frac{{(cx + d)(a) - (ax + b)(c)}}{{{{(cx + d)}^2}}}$
$ = \frac{{acx + ad - acx - bc}}{{{{(cx + d)}^2}}} = \frac{{ad - bc}}{{{{(cx + d)}^2}}}$
View full question & answer→Question 62 Marks
Find the derivative of function $(ax + b)(cx + d)^2$ (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).
AnswerHere $f(x) = (ax + b) (cx + d)^2$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}[(ax + b){(cx + d)^2}]$
= $(ax + b ) \frac{d}{{dx}}{(cx + d)^2} + {(cx + d)^2} \cdot \frac{d}{{dx}}(ax + b)$
= $(ax + b) \times 2(cx + d) \times c + (cx + d)^2 \times a$
= $2c(ax + b)(cx + d) + a(cx + d)^2$
View full question & answer→Question 72 Marks
Find the derivative of function $\frac{x}{{{{\sin }^n}x}}$ (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).
AnswerHere $f(x) = \frac{x}{{{{\sin }^n}x}}$
$\therefore \;f'(x) = \frac{d}{{dx}}\left[ {\frac{x}{{{{\sin }^n}x}}} \right]$
$ = \frac{{{{\sin }^n}x\frac{d}{{dx}}(x) - x\frac{d}{{dx}}({{\sin }^n}x)}}{{{{({{\sin }^n}x)}^2}}}$
$= \frac{{{{\sin }^n}x \times 1 - x.n{{\sin }^{n - 1}}x\frac{d}{{dx}}(\sin x)}}{{{{({{\sin }^n}x)}^2}}}$
$= \frac{{{{\sin }^n}x - nx{{\sin }^{n - 1}}x\cos x}}{{{{({{\sin }^n}x)}^2}}}$
$ = \frac{{{{\sin }^{n - 1}}x[\sin x - nx\cos x]}}{{{{\sin }^{2x}}x}}$
$ = \frac{{\sin x - nx\cos x}}{{{{\sin }^{n + 1}}x}}$
View full question & answer→Question 82 Marks
Find the derivative of function (px + q)$\left( {\frac{r}{x} + s} \right)$(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).
AnswerHere f (x) = (px + q)$\left( {\frac{r}{x} + s} \right)$
$\therefore f{\text{'}}(x) = \frac{d}{{dx}}[(px + q) \left( {\frac{r}{x} + s} \right) ]$
$=\frac d{dx}\left(pr+psx\;+\frac{qr}x+\;sq\right)$
$=\;\frac d{dx}\left(pr\right)\;+\frac d{dx}\left(psx\right)\;+\frac d{dx}\left(\frac{qr}x\right)\;+\frac d{dx}\left(sq\right)$
$=\;0\;+ps(\;1)\;+qr\;\left(\frac{-1}{x^2}\right)\;+0\\=ps\;-\;\frac{qr}{x^2}$
View full question & answer→Question 92 Marks
Find the derivative of the function $\frac{x^{2} \cos \left(\frac{\pi}{4}\right)}{\sin x}$ (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)
AnswerLet f(x) = $\frac{x^{2} \cos \left(\frac{\pi}{4}\right)}{\sin x}$
By quotient rule,we have,
$f^{\prime}(x)=\cos \frac{\pi}{4} \cdot\left[\frac{\sin x \frac{d}{d x}\left(x^{2}\right)-x^{2} \frac{d}{d x}(\sin x)}{\sin ^{2} x}\right]$
$=\cos \frac{\pi}{4} \cdot\left[\frac{\sin x \times 2 x-x^{2} \cos x}{\sin ^{2} x}\right]$
$\therefore \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x} \cos \frac{\pi}{4}[2 \sin \mathrm{x}-\mathrm{x} \cos \mathrm{x}]}{\sin ^{2} \mathrm{x}}$
View full question & answer→Question 102 Marks
Find the derivative of the function $f(x) = \frac{{4x + 5\sin x}}{{3x + 7\cos x}}$
AnswerHere $f(x) = \frac{{4x + 5\sin x}}{{3x + 7\cos x}}$
$\therefore\;f'(x)=\frac{(3x+7\cos x)\frac d{dx}(4x+5\sin x)-(4x+5\sin x)\frac d{dx}(3x+7\cos x)}{{(3x+7\cos x)}^2}$
$ = \frac{{(3x + 7\cos x)(4+ 5\cos x) - (4x + 5\sin x)(3 - 7\sin x)}}{{{{(3x + 7\cos x)}^2}}}$
$=\;\frac{12x+15x\cos x+28\cos x+35\cos^2x-12x+28x\sin x-15\sin x+35\;\sin^2x}{{(3x+7\cos x)}^2}$
$ = \frac{{15x\cos x + 28\cos x + 28x\sin x - 15\sin x + 35({{\cos }^2}x + {{\sin }^2}x)}}{{{{(3x + 7\cos x)}^2}}}$
$ = \frac{{15x\cos x + 28\cos x + 28x\sin x - 15\sin x + 35}}{{{{(3x + 7\cos x)}^2}}}$
View full question & answer→Question 112 Marks
Find the derivative of function $\frac{{a + b\sin x}}{{c + d\cos x}}$ (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)
AnswerHere $f(x) = \frac{{a + b\sin x}}{{c + d\cos x}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{a + b\sin x}}{{c + d\cos x}}} \right]$
$= \frac{{(c + d\cos x)\frac{d}{{dx}}(a + b\sin x) - (a + b\sin x)\frac{d}{{dx}}(c + d\cos x)}}{{{{(c + d\cos )}^2}}}$
$= \frac{{(c + d\cos x)(b\cos x) - (a + b\sin x)( - d\sin x)}}{{{{(c + d\cos x)}^2}}}$
$= \frac{{bc\cos x + bd{{\cos }^2}x + ad\sin x + bd{{\sin }^2}x}}{{{{(c + d\cos x)}^2}}}$
$ = \frac{{bc\cos x + ad\sin x + bd({{\cos }^2}x + {{\sin }^2}x)}}{{{{(c + d\cos x)}^2}}}$
$ = \frac{{bc\cos x + ad\sin x + bd}}{{{{(c + d\cos x)}^2}}}$
View full question & answer→Question 122 Marks
Find the derivative of function (x + a) (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).
AnswerHere f(x) = x + a
$\therefore {\text{f'}}(x) = \frac{d}{{dx}}(x + a) = 1$
View full question & answer→Question 132 Marks
Find the derivative of function $\frac{{\sec x - 1}}{{\sec x + 1}}$ (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)
Answer$f(x) = \frac{{\sec x - 1}}{{\sec x + 1}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{\sec x - 1}}{{\sec x + 1}}} \right]$
$= \frac{{(\sec + 1)\frac{d}{{dx}}(\sec x - 1) - (\sec x - 1)\frac{d}{{dx}}(\sec x + 1)}}{{{{(\sec x + 1)}^2}}}$
$= \frac{{(\sec x + 1)(\sec x\tan x) - (\sec x - 1)(\sec x\tan x)}}{{{{(\sec x + 1)}^2}}}$
$ = \frac{{{{\sec }^2 x}\tan x + \sec x\tan x - {{\sec }^2}x\tan x + \sec x\tan x}}{{{{(\sec x + 1)}^2}}}$
$= \frac{{2\sec x\tan x}}{{{{(\sec x + 1)}^2}}}$
View full question & answer→Question 142 Marks
Find the derivative of function $\frac{{\sin x + \cos x}}{{\sin x - \cos x}}$ (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).
AnswerHere $f(x) = \frac{{\sin x + \cos x}}{{\sin x - \cos x}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right]$
$= \frac{{(\sin x - \cos x)\frac{d}{{dx}}(\sin x + \cos x) - (\sin x + \cos x)\frac{d}{{dx}}(\sin x - \cos x)}}{{{{(\sin x - \cos x)}^2}}}$
$= \frac{{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}}{{{{(\sin x - \cos x)}^2}}}$
$= \frac{{ - {{(\sin x - \cos x)}^2} - {{(\sin x + \cos x)}^2}}}{{{{(\sin x - \cos x)}^2}}}$
$= \frac{{ - ({{\sin }^2}x - {{\cos }^2}x + 2\sin x\cos x - {{\sin }^2}x - {{\cos }^2}x - {{\cos }^2}x - 2\sin x\cos x}}{{{{(\sin x - \cos x)}^2}}}$
$= \frac{{ - 2({{\sin }^2}x + {{\cos }^2}x)}}{{{{(\sin x - \cos x)}^2}}}$$= \frac{{ - 2}}{{{{(\sin x - \cos x)}^2}}}$
View full question & answer→Question 152 Marks
Find the derivative of function $f(x) = \frac{{\cos x}}{{1 + \sin x}}$ (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).
AnswerHere $f(x) = \frac{{\cos x}}{{1 + \sin x}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{\cos x}}{{1 + \sin x}}} \right]$
$= \frac{{(1 + \sin x)\frac{d}{{dx}}(\cos x) - \cos x \cdot \frac{d}{{dx}}(1 + \sin x)}}{{{{(1 + \sin x)}^2}}}$
$= \frac{{(1 + \sin x)( - \sin x) - \cos x(\cos x)}}{{{{(1 + \sin x)}^2}}}$
$ = \frac{{ - \sin x - {{\sin }^2}x - {{\cos }^2}x}}{{{{(1 + \sin x)}^2}}}$$= \frac{{ - \sin x - ({{\sin }^2}x + {{\cos }^2}x)}}{{{{(1 + \sin x)}^2}}}$
$= \frac{{ - \sin x - 1}}{{{{(1 + \sin x)}^2}}} = \frac{{ - (1 + \sin x)}}{{{{(1 + \sin x)}^2}}} = \frac{{ - 1}}{{1 + \sin x}}$
View full question & answer→Question 162 Marks
Find the derivative of function sin (x + a) (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).
AnswerHere f (x) = sin (x + a)
$\therefore$ f'(x) = $\frac{d}{{dx}}$ [sin (x + a)]
= cos (x + a). $\frac{d}{{dx}}$(x + a)
= cos (x + a)
View full question & answer→Question 172 Marks
Find the derivative of function $(ax + b)^n (cx + d)^m$ (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).
Answer$\text { Here } f(\mathrm{x})=(\mathrm{ax}+\mathrm{b})^{\mathrm{n}}(\mathrm{cx}+\mathrm{d})^{\mathrm{m}}$
$\therefore f^{\prime}(x)=\frac{d}{d x}\left[(a x+b)^n(c x+d)^m\right]$
$=(a x+b)^n \frac{d}{d x}(c x+d)^m+(c x+d)^m \frac{d}{d x}(a x+b)^n$
$=(\mathrm{ax}+\mathrm{b})^{\mathrm{n}} \cdot \mathrm{m}(\mathrm{cx}+\mathrm{d})^{\mathrm{m}-1} \cdot \frac{d}{d x}(\mathrm{cx}+\mathrm{d})+(\mathrm{cx}+\mathrm{d})^m \mathrm{n}(a x+b)^{\mathrm{n}-1} \cdot \frac{d}{d x}(a x+b)$
$=\mathrm{cm}(\mathrm{ax}+\mathrm{b})^{\mathrm{n}}(\mathrm{cx}+\mathrm{d})^{\mathrm{m}-1}+\mathrm{an}(\mathrm{cx}+\mathrm{d})^{\mathrm{m}}(\mathrm{ax}+\mathrm{b})^{\mathrm{n}-1}$
$=(\mathrm{ax}+\mathrm{b})^{\mathrm{n}-1}(\mathrm{cx}+\mathrm{d})^{\mathrm{m}-1}[\mathrm{~cm}(a x+b)+\mathrm{an}(\mathrm{cx}+\mathrm{d})]$
View full question & answer→Question 182 Marks
Find the derivative of function $4\sqrt x - 2$ (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).
AnswerHere $f(x) = 4\sqrt x - 2$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {4\sqrt x - 2} \right]$
$= 4\frac{d}{{dx}}(\sqrt x ) - \frac{d}{{dx}}(2)$
$= 4 \times \frac{1}{{2\sqrt x }} - 0 = \frac{2}{{\sqrt x }}$
View full question & answer→Question 192 Marks
Find the derivative of function $\frac{a}{{{x^4}}} - \frac{b}{{{x^2}}} + \cos x$ (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).
AnswerHere $f(x) = \frac{a}{{{x^4}}} - \frac{b}{{{x^2}}} + \cos x = a{x^-4} - b{x^{ - 2}} + \cos x$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}[a{x^{ - 4}} - b{x^2} + \cos x]$$= a\frac{d}{{dx}}({x^{ - 4}}) - b\frac{d}{{dx}}({x^{ - 2}}) + \frac{d}{{dx}}(\cos x)$
$ - a{x^{ - 5}} + 2b{x^{ - 3}} - \sin x = \frac{{ - 4a}}{{{x^5}}} + \frac{{2b}}{{{x^3}}} - \sin x$
$- 4a{x^{ - 5}} + 2b{x^{ - 3}} - \sin x = \frac{{ - 4a}}{{{x^5}}} + \frac{{2b}}{{{x^3}}} - \sin x$
View full question & answer→Question 202 Marks
Find the derivative of $\frac{{{x^n} - {a^n}}}{{x - a}}$ for some constant a.
AnswerHere $f(x)\frac{{{x^n} - {a^n}}}{{x - a}}$
$\therefore \;f(x) = \frac{d}{{dx}}\left[ {\frac{{{x^n} - {a^n}}}{{x - a}}} \right]$
$= \frac{{(x - a)\frac{d}{{dx}}({x^n} - {a^n}) - ({x^n} - {a^n})\frac{d}{{dx}}(x - a)}}{{{{(x - a)}^2}}}$
$ = \frac{{(x - a) \times n{x^{n - 1}} - ({x^n} - {a^n}) \times 1}}{{{{(x - a)}^2}}}$
$= \frac{{n{x^n} - an{x^{n - 1}} - {x^n} + {a^n}}}{{{{(x - a)}^2}}}$
View full question & answer→Question 212 Marks
For some constants, a and b, find the derivative of f(x) $= \frac{{x - a}}{{x - b}}$
AnswerHere f (x) $= \frac{{x - a}}{{x - b}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left( {\frac{{x - a}}{{x - b}}} \right)$
$ = \frac{{(x - b)\frac{d}{{dx}}(x - a) - (x - a)\frac{d}{{dx}}(x - b)}}{{{{(x - b)}^2}}}$
$ = \frac{{(x - b) \times 1 - (x - a) \times 1}}{{{{(x - b)}^2}}} = \frac{{x - b - x + a}}{{{{(x - b)}^2}}}$
$= \frac{{a - b}}{{{{(x - b)}^2}}}$
View full question & answer→Question 222 Marks
For some constants a and b, find the derivative of (x - a)(x - b)
AnswerHere f (x) = (x - a)(x - b)
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}(x - a)(x - b)$
$ = (x - a)\frac{d}{{dx}}(x - b) + (x - b)\frac{d}{{dx}}(x - a)$
= (x - a) × 1 + (x - b) × 1
= x - a + x - b = 2x - a - b
View full question & answer→Question 232 Marks
Find the derivative of $x^n+a x^{n-1}+a^2 x^{n-2}+\ldots+a^{n-1} x+a^n$ for some fixed real number $a$.
AnswerLet $f(x)=x^n+a x^{n-1}+a^2 x^{n-2}+\ldots .+a^{n-1} x+a^n$
On differentiating both sides, we get
$f^{\prime}(x)=n x^{n-1}+a(n-1) x^{n-2}+a^2(n-2) x^{n-3}+\ldots . .+a^{n-1} \cdot 1+0$
On putting $x=a$ both sides, we get
$f^{\prime}(a)=n n^{n-1}+a(n-1) a^{n-2}+a^2(n-2) a^{n-3}+\ldots+a^{n-1}$
$=n a^{n-1}+(n-1) a^{n-1}+(n-2) a^{n-1}+\ldots .+a^{n-1}$
$=a^{n-1}[n+(n-1)+(n-2)+\ldots .+1]$
$\left[\because\right.$ sum of $n$ natural numbers $\left.=\frac{n(n+1)}{2}\right]$
$\mathrm{f}^{\prime}(\mathrm{a})=\frac{n(n+1)}{2} a^{n-1}$
View full question & answer→Question 242 Marks
For the function $f(x) = \frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + ... + \frac{{{x^2}}}{2} + x + 1$ prove that f'(1) = 100f'(0)
AnswerHere $f(x) = \frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + ... + \frac{{{x^2}}}{2} + x + 1$
$f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + ... + \frac{{{x^2}}}{2} + x + 1} \right]$
$= \frac{1}{{100}}\frac{d}{{dx}}({x^{100}}) + \frac{1}{{99}}\frac{d}{{dx}}({x^{99}}) + ... + \frac{1}{2}\frac{d}{{dx}}({x^2}) + \frac{d}{{dx}}(x) + \frac{d}{{dx}}(1)$
$= \frac{1}{{100}} \times 100{x^{99}} + \frac{1}{{99}} \times 99{x^{98}} + ... + \frac{1}{2} \times 2x + 1 + 0$
$=x^{99}+x^{98}+\ldots+x+1$
$\text { Now } f^{\prime}(1)=(1)^{99}+(1)^{98}+\ldots+(1)+1=100$
$f^{\prime}(0)=(0)^{99}+(0)^{98}+\ldots+0+1=1$
Which shows that $f^{\prime}(1)=100 f^{\prime}(0)$
View full question & answer→Question 252 Marks
Find the derivative of $\left( \frac { x + 1 } { x - 1 } \right)$ from the first principle.
AnswerWe have, $f ( x ) = \frac { x + 1 } { x - 1 }$
By first principle of derivative, we have
${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\frac{{(x + h) + 1}}{{(x + h) - 1}} - \frac{{x + 1}}{{x - 1}}} \right]}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{(x + h + 1)(x - 1) - (x + 1)(x + h - 1)}}{{h(x + h - 1)(x - 1)}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{x^2} + xh - h - 1} \right) - \left( {{x^2} + xh + h - 1} \right)}}{{h(x + h - 1)(x - 1)}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2h}}{{h(x + h - 1)(x - 1)}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2}}{{(x + h - 1)(x - 1)}} = \frac{{ - 2}}{{{{(x - 1)}^2}}}$
View full question & answer→Question 262 Marks
Find the derivative of $1/x^2$ from the first principle.
AnswerHere ${\text{f}}(x) = \frac{1}{{{x^2}}}$
Then ${\text{f}}(x + h) = \frac{1}{{{{(x + h)}^2}}}$
We know that ${\text{f'}}(x) = \mathop {\lim }\limits_{x \to 0} \frac{{f(x + h) - f(x)}}{h}$
$\Rightarrow \;{\text{f'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{{{(x + h)}^2}}} - \frac{1}{{{x^2}}}}}{h}$$= \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} - {{(x + h)}^2}}}{{h{x^2}{{(x + h)}^2}}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} - {x^2} - {h^2} - 2xh}}{{h{x^2}{{(x + h)}^2}}} = \mathop {\lim }\limits_{h \to 0} \frac{{h( - h - 2x)}}{{h{x^2}{{(x + h)}^2}}}$
$ = \frac{{ - 2x}}{{{x^2} \times {x^2}}} = \frac{{ - 2}}{{{x^3}}}$
View full question & answer→Question 272 Marks
Find the derivative of (x -1) (x - 2) from first principle.
AnswerWe have, $f(x) = (x - 1)(x - 2)$
$= x^2 - 3x + 2$
By first principle of derivative, we have
${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{(x + h)}^2} - 3(x + h) + 2} \right] - \left[ {{x^2} - 3x + 2} \right]}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\left( {{x^2} + {h^2} + 2xh - 3x - 3h + 2} \right] - \left[ {{x^2} - 3x + 2} \right]} \right.}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{2hx + {h^2} - 3h}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{h(2x + h - 3)}}{h}$
$= 2 x - 3$
View full question & answer→Question 282 Marks
Find the derivative of $(x^3 - 27)$ from first principle.
AnswerWe have, $f(x) = x^3 - 27$
By using first principle of derivative,
${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$
$\therefore \quad {f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{(x + h)}^3} - 27} \right] - \left[ {{x^3} - 27} \right]}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{{x^3} + 3{x^2}h + 3{h^2}x + {h^3} - 27 - {x^3} + 27}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{3{x^2}h + 3x{h^2} + {h^3}}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} (3x^2 + 3xh + h^2) = 3x^2$
View full question & answer→Question 292 Marks
Find the derivative of 99x at x = 100
AnswerHere $\frac{d}{{dx}}(99x) = 99$
$\therefore $ Derivative of 99x at x = 100 = 99
View full question & answer→Question 302 Marks
Find the derivative of the function $5 \sin x-6 \cos x+7$
AnswerLet f(x) = 5 sin x – 6 cos x + 7
Therefore, we have,
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}(5 \sin \mathrm{x}-6 \cos \mathrm{x}+7)$
$=5 \frac{d}{d x}(\sin x)-6 \frac{d}{d x}(\cos x)+\frac{d}{d x}(7)$
= 5 $\times$ cos x – 6 $\times$ (- sin x) + 0
$\therefore$ f’(x) = 5 cos x + 6 sin x
View full question & answer→Question 312 Marks
Find the derivative of function 3 cot x + 5 cosec x
AnswerHere f (x) = 3 cot x + 5 cosec x
$\therefore \;{\text{f'}}(x) = \frac{d}{{dx}}[3\cot x + 5c{\text{osec }}x]$
$= 3\frac{d}{{dx}}(\cot x) + 5\frac{d}{{dx}}(cosecx)$
$=-3 \operatorname{cosec}^2 x-5 \operatorname{cosec} x \cot x$
View full question & answer→Question 322 Marks
Find the derivative of function sec x.
AnswerHere f (x) = sec x
$\therefore \;{\text{f'}}(x) = \frac{d}{{dx}}(\sec x)$
= sec x tan x
View full question & answer→Question 332 Marks
Find the derivative of function sin x cos x.
AnswerHere f (x) = sin x cos x
$\therefore \;f{\text{}}(x) = \frac{d}{{dx}}(\sin x\cos x)$
$= \sin x\frac{d}{{dx}}(\cos x) + \cos x\frac{d}{{dx}}(\sin x)$
$=\sin x(-\sin x)+\cos x(\cos x)=\cos ^2 x-\sin ^2 x=\cos 2 x$
View full question & answer→Question 342 Marks
Find the derivative of cos x from first principle.
AnswerHere f (x) = cos x
Then f (x + h) = cos (x + h)
We know that $f{\text{'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{(x + h) - f(x)}}{h}$
$\Rightarrow f{\text{'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\cos (x + h) - \cos x}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2\sin \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{h}{2}} \right)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} - \sin \left( {\frac{{2x + h}}{2}} \right).\frac{{\sin \left( {\frac{h}{2}} \right)}}{{\frac{h}{2}}}$
= -sin x
View full question & answer→Question 352 Marks
Find the derivative of $x^2 - 2$ at $x = 10$
AnswerHere $\frac{d}{{dx}}({x^2} - 2)$ = 2x - 0 = 2x
$\therefore$ Derivative of $x^2 - 2$ at $x = 10 = 2 \times 10= 20$
View full question & answer→Question 362 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{ax + b}}{{cx + 1}}$
AnswerHere $\mathop {\lim }\limits_{x \to 0} \frac{{ax + b}}{{cx + 1}} = \frac{{a \times 0 + b}}{{c \times 0 + 1}} = \frac{b}{1} = b$
View full question & answer→Question 372 Marks
Evaluate $\mathop {\lim }\limits_{x \to 3} \frac{{{x^4} - 81}}{{2{x^2} - 5x - 3}}$
AnswerHere $\mathop {\lim }\limits_{x \to 3} \frac{{{x^4} - 81}}{{2{x^2} - 5x - 3}}\left[ {\frac{0}{0}{\text{from}}} \right]$
$= \mathop {\lim }\limits_{x \to 3} \frac{{({x^2} + 9)(x + 3)(x - 3)}}{{(x - 3)(2x + 1)}}$
$= \mathop {\lim }\limits_{x \to 3} \frac{{({x^2} + 9)(x + 3)}}{{(2x + 1)}} = \frac{{({3^2} + 9)(3 + 3)}}{{(2 \times 3 + 1)}} = \frac{{108}}{7}$
View full question & answer→Question 382 Marks
Evaluate $\mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} - x - 10}}{{{x^2} - 4}}$
AnswerHere $\mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} - x - 10}}{{{x^2} - 4}}\left[ {\frac{0}{0}{\text{from}}} \right]$
$= \mathop {\lim }\limits_{x \to 2} \frac{{(x - 2)(3x + 5)}}{{(x + 2)(x - 2)}}$
$= \mathop {\lim }\limits_{x \to 2} \frac{{3x - 5}}{{x + 2}} = \frac{{6 + 5}}{{2 + 2}} = \frac{{11}}{4}$
View full question & answer→Question 392 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{{{(X + 1)}^5} - 1}}{x}$
AnswerHere $\mathop {\lim }\limits_{x \to 0} \frac{{{{(X + 1)}^5} - 1}}{x}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{{{(X + 1)}^5} - 1}}{{(x + 1) - 1}}$
Putting x + 1 = y, as $x \to 0,\;y \to 1$
$\therefore \;\mathop {\lim }\limits_{y \to 0} \frac{{{y^5} - 1}}{{y - 1}} = 5.{(1)^{5 - 1}}$
$= 5 \times 1 = 5\left[ {\because \;\mathop {\lim }\limits_{x \to a} \frac{{{x^n} - {a^n}}}{{x - a}} = n \cdot {a^{n - 1}}} \right]$
View full question & answer→Question 402 Marks
Evaluate $\mathop {\lim }\limits_{x \to - 1} \frac{{{x^{10}} + {x^5} + 1}}{{x - 1}}$
AnswerHere $\mathop {\lim }\limits_{x \to - 1} \frac{{{x^{10}} + {x^5} + 1}}{{x - 1}}$
$ = \frac{{{{( - 1)}^{10}} + {{( - 1)}^5} + 1}}{{ - 1 - 1}} = \frac{{1 - 1 + 1}}{{ - 2}} = \frac{{ - 1}}{2}$
View full question & answer→Question 412 Marks
Evaluate $\mathop {\lim }\limits_{x \to 4} \frac{{4x + 3}}{{x - 2}}$
AnswerHere $\mathop {\lim }\limits_{x \to 4} \frac{{4x + 3}}{{x - 2}} = \frac{{4 \times 4 + 3}}{{4 - 2}} = \frac{{19}}{2}$
View full question & answer→Question 422 Marks
If $f(x) = \left\{ {\begin{array}{*{20}{c}} {m{x^2} + n,} \\ {nx + m,} \\ {n{x^3} + m,} \end{array}} \right.\begin{array}{*{20}{c}} {x < 0} \\ {0 \le x \le 1} \\ {x > 1} \end{array}$ .
For what integers m and n does both $\mathop {\lim }\limits_{x \to 0} f(x)$ and $\mathop {\lim }\limits_{x \to 1} f(x)$ exist?
AnswerIt is given that
$\mathop {\lim }\limits_{x \to 0}$ f(x) and $\mathop {\lim }\limits_{x \to 1}$ f(x) both exist.
$\Rightarrow \;\mathop {\lim }\limits_{x \to {0^ - }}$ f(x) and = $\mathop {\lim }\limits_{x \to {0^ +}}$f(x) = $\mathop {\lim }\limits_{x \to {1^ + }}$ f(x)
Now $\mathop {\lim }\limits_{x \to {0^- }}$ f(x) = $\mathop {\lim }\limits_{x \to {0 }} (mx^2+ n) = n$
$\mathop {\lim }\limits_{x \to {0^ + }}$ f(x) = $\mathop {\lim }\limits_{x \to {0 }}$ (nx + m) = m
Now $\mathop {\lim }\limits_{x \to {0^ + }}$ f(x) = $\mathop {\lim }\limits_{x \to {0^ + }}$ f(x) $\Rightarrow$ n = m ...(i)
For $\mathop {\lim }\limits_{x \to 0}$ f(x) to exist we need m = n
Also, $\mathop {\lim }\limits_{x \to {1^ - }}$ f(x) = $\mathop {\lim }\limits_{x \to {1}} (nx + m) = n + m$
$\mathop {\lim }\limits_{x \to {1^ + }}$ f(x) = $\mathop {\lim }\limits_{x \to {1}} (nx^3 + m) = n + m$
Now $\mathop {\lim }\limits_{x \to {1^ - }}$ f(x) = $\mathop {\lim }\limits_{x \to {1^ + }}$ f(x) $\Rightarrow$ n + m = n + m
Thus $\mathop {\lim }\limits_{x \to 0}$ f(x) exists for any integral value of m and n.
View full question & answer→Question 432 Marks
If the function f(x) satisfies $\mathop {\lim }\limits_{x \to 1} \frac{{f(x) - 2}}{{{x^2} - 1}} = \pi ,$ then evaluate $\mathop {\lim }\limits_{x \to 1} f(x)$.
AnswerGiven, $\mathop {\lim }\limits_{x \to 1} \frac{{f(x) - 2}}{{{x^2} - 1}} = \pi \Rightarrow \frac{{\mathop {\lim }\limits_{x \to 1} [f(x) - 2]}}{{\mathop {\lim }\limits_{x \to 1} \left( {{x^2} - 1} \right)}} = \pi $
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} [f(x) - 2] = \pi \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - 1} \right)$
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} f(x) - 2 = \pi \left( {{1^2} - 1} \right)$
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} f(x) - 2 = \pi \times 0 \Rightarrow \mathop {\lim }\limits_{x \to 1} f(x) - 2 = 0$
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} f(x) = 2$
View full question & answer→Question 442 Marks
If f(x) = $ \left\{ {\begin{array}{*{20}{c}} {|x| + 1} \\ 0 \\ {|x| - 1} \end{array}} \right.\begin{array}{*{20}{c}} {x < 0} \\ {x = 0} \\ {x > 0} \end{array}$ for what values of a does $\mathop {\lim }\limits_{x \to a}$ f(x) exists?
AnswerHere f(x) = $ \left\{ {\begin{array}{*{20}{c}} {|x| + 1} \\ 0 \\ {|x| - 1} \end{array}} \right.\begin{array}{*{20}{c}} {x < 0} \\ {x = 0} \\ {x > 0} \end{array}$
$\Rightarrow$ f(x) = $\left\{ {\begin{array}{*{20}{c}} { - x + 1,} \\ {0,} \\ {x - 1,} \end{array}} \right.\begin{array}{*{20}{c}} {x < 0} \\ {x = 0} \\ {x > 0} \end{array}$
$\therefore \;\mathop {\lim }\limits_{x \to a}$ f(x) exists for all a $\ne$ 0
Now we see that $\mathop {\lim }\limits_{x \to 0}$ f(x) exist or not
L.H.L.= $\mathop {\lim }\limits_{x \to {0^ - }}$ f(x) = $\mathop {\lim }\limits_{x \to {0^ - }}$ (-x + 1) = 1
R.H.L =$\mathop {\lim }\limits_{x \to {0^ - }}$ f(x) = $\mathop {\lim }\limits_{x \to {0 }}$ (x - 1) = -1
$\therefore$ L.H.L at x = 0 $\ne$ R.H.L as x 0
Thus $\therefore \;\mathop {\lim }\limits_{x \to 0}$ f(x) does not exist.
View full question & answer→Question 452 Marks
Let $a_1, a_2, ....,a_n$ be fixed real numbers and define a function $f(x) = (x - a_1)(x - a_2)... (x - a_n)$. What is & $\mathop {\lim }\limits_{x \to {a_1}} f(x)$? For some $a \ne {a_1},\;{a_2}...{a_n},$ compute $\mathop {\lim }\limits_{x \to {a}} f(x)$
AnswerHere $f(x)=\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)$
Now $\mathop {\lim }\limits_{x \to {a_1}} f(x) = \mathop {\lim }\limits_{x \to {a_1}} (x - {a_1})(x - {a_2})...(x - {a_n})$
$=\left(a_1-a_1\right)\left(a_1-a_2\right) \ldots\left(a_1-a_n\right)$
$=0 \times\left(a_1-a_2\right) \ldots\left(a_1-a_n\right)=0$
Also $\mathop {\lim }\limits_{x \to a} f(x) = \mathop {\lim }\limits_{x \to a} (x - {a_1})(x - {a_2})...(x - {a_n})$
$=\left(a-a_1\right)\left(a-a_2\right) \ldots\left(a-a_n\right)$
View full question & answer→Question 462 Marks
Suppose f(x) = $ \left\{ {\begin{array}{*{20}{c}} {a + bx,} \\ {4,} \\ {b - ax,} \end{array}} \right.\begin{array}{*{20}{c}} {x < 1} \\ {x = 1} \\ {x > 1} \end{array}$ and if $\mathop {\lim }\limits_{x \to 1}$ f(x) = f(1) what are possible values of a and b?
AnswerHere f(x) = $ \left\{ {\begin{array}{*{20}{c}} {a + bx,} \\ {4,} \\ {b - ax,} \end{array}} \right.\begin{array}{*{20}{c}} {x < 1} \\ {x = 1} \\ {x > 1} \end{array}$
Also $\mathop {\lim }\limits_{x \to 1}$ f(x) = f(1)
$\Rightarrow \mathop {\lim }\limits_{x \to 1}$ f(x) = $\mathop {\lim }\limits_{x \to {1^ + }}$ f(x) = f(1) = 4
$\Rightarrow \mathop {\lim }\limits_{x \to {1^ - }}$ f(x) = 4 and $\mathop {\lim }\limits_{x \to {1^ + }}$ f(x) = 4
Now $\mathop {\lim }\limits_{x \to {1^ - }}$ f(x) = $\mathop {\lim }\limits_{x \to {1^ - }}$ (a + bx)
Put x = 1 - h as x $\to$ 1, h $\to$ 0
$\therefore \;\mathop {\lim }\limits_{h \to 0}$ [a + b(1 - h)] = $\mathop {\lim }\limits_{h \to 0}$ [a + b - bh] = a + b
Also $\mathop {\lim }\limits_{x \to {1^ + }}$ f(x) = $\mathop {\lim }\limits_{x \to {1^ + }}$ (b - ax)
Put x = 1 + h as x $\to$ 1, h $\to$ 0
$\therefore \;\mathop {\lim }\limits_{h \to 0}$ [a - b(1 + h)] = $\mathop {\lim }\limits_{h \to 0}$ [b - a - ah] = b - a
Putting values from (ii) and (iii) in (i)
$\therefore $ a + b = 4 and -a + b = 4
Solving these equations, we have
a = 0 and b = 4
View full question & answer→Question 472 Marks
Find $\mathop {\lim }\limits_{x \to 5} f(x)$, where f(x) = |x| - 5
AnswerHere f(x) = |x| - 5
L.H.L. $\mathop {\lim }\limits_{x \to 5} f(x) = \mathop {\lim }\limits_{x \to {5^ - }} |x| - 5$
Put x = 5 – h as x → 5, h → 0
$\therefore \;\mathop {\lim }\limits_{h \to 0} \left| {5 - h} \right| - 5 = \mathop {\lim }\limits_{h \to 0} 5 - h - 5 = \mathop {\lim }\limits_{h \to 0} ( - h) = 0$
R.H.L. $= \mathop {\lim }\limits_{x \to {5^ + }} f(x) = \mathop {\lim }\limits_{x \to {5^ + }} |x| - 5$
Put x = 5 + h as x → 5, h →0
$\therefore \;\mathop {\lim }\limits_{h \to 0} \left| {5 + h} \right| - 5 = \mathop {\lim }\limits_{h \to 0} 5 + h - 5 = \mathop {\lim }\limits_{h \to 0} h = 0$
Now L.H.L. = R.H.L
Thus limit exists at x = 5 and $\mathop {\lim }\limits_{x \to 0} f(x) = 0$
View full question & answer→Question 482 Marks
Find $\mathop {\lim }\limits_{x \to 0} f(x)$ where $f ( x ) = \left\{ \begin{array} { c c } { \frac { x } { | x | } , } & { x \neq 0 } \\ { 0 , } & { x = 0 } \end{array} \right.$
AnswerHere $f ( x ) = \left\{ \begin{array} { c c } { \frac { x } { | x | } , } & { x \neq 0 } \\ { 0 , } & { x = 0 } \end{array} \right.$
L.H.L. $ = \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} \frac{x}{{|x|}}$
Put x = 0 – h as x → 0, h → 0
$\therefore \mathop {\lim }\limits_{h \to 0} \frac{{0 - h}}{{|0 - h|}} = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{| - h|}} = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{h} = - 1$
R.H.L. $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{|x|}}$
Put x = 0 + h as $\mathrm { x } \rightarrow 0 , \mathrm { h } \rightarrow 0$
$\therefore \mathop {\lim }\limits_{x \to 0} \frac{{0 + h}}{{|0 + h|}} = \mathop {\lim }\limits_{n \to 0} \frac{h}{{|h|}} = \mathop {\lim }\limits_{h \to 0} \frac{h}{h} = 1$
Now L.H.L. $\neq$ R.H.L.
Thus limit does not exist at x = 0.
View full question & answer→Question 492 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0}$ f(x), where f(x) = $\left\{ \begin{array} { l l } { \frac { | x | } { x } , } & { x \neq 0 } \\ { 0 , } & { x = 0 } \end{array} \right.$
AnswerWe have, f(x) = $\left\{ {\begin{array}{*{20}{l}} {\frac{{|x|}}{x},}&{x \ne 0} \\ {0,}&{x = 0} \end{array}} \right.$
LHL = $\mathop {\lim }\limits_{x \to {0^ - }}$ f(x) = $\mathop {\lim }\limits_{x \to {0^ - }} \frac{{|x|}}{x} = \mathop {\lim }\limits_{h \to 0} \frac{{|0 - h|}}{{(0 - h)}}$ [putting x = 0 - h as x $\rightarrow$ 0, then h $\rightarrow$ 0]
= $\mathop {\lim }\limits_{h \to 0} \frac{{| - h|}}{{ - h}}$
= $\mathop {\lim }\limits_{h \to 0} \frac{{ h}}{{ - h}}$ [$\because$ |-x| = x]
= - 1
RHL = $\mathop {\lim }\limits_{x \to {0^ + }}$ f(x) = $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{|x|}}{x} = \mathop {\lim }\limits_{h \to 0} \frac{{|0 + h|}}{{(0 + h)}} = \mathop {\lim }\limits_{h \to 0} \frac{h}{h}$ = 1 [putting x = 0 + h as x $\rightarrow$ 0, then h $\rightarrow$ 0]
$\mathop {\lim }\limits_{h \to {0^ - }}$ f(x) $\ne \mathop {\lim }\limits_{x \to {0^ + }}$ f(x)
Limit does not exist at x = 0
View full question & answer→Question 502 Marks
Find $\mathop {\lim }\limits_{x \to 1} f(x)$ where f(x) = $\left\{ {\begin{array}{*{20}{c}} {{x^2} - 1,}&{x \le 0} \\ { - {x^2} - 1,}&{x > 1} \end{array}} \right.$
AnswerHere $\mathop {\lim }\limits_{x \to 1} f(x)$ = $\left\{ {\begin{array}{*{20}{c}} {{x^2} - 1,}&{x \le 0} \\ { - {x^2} - 1,}&{x > 1} \end{array}} \right.$
L.H.L. $\mathop {\lim }\limits_{x \to 1} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} ({x^2} - 1)$
Put x = 1 - h as $x \to 1,\;h \to 0$
$\therefore \;\mathop {\lim }\limits_{h \to 0} [{(1 + h)^2} - 1] = \mathop {\lim }\limits_{h \to 0} [1 + {h^2} - 1]$
R.H.L. $= \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} ( - {x^2} - 1)$
Put x = 1 + h as $x \to 1,\;h \to 0$
$\therefore \;\mathop {\lim }\limits_{h \to 0} [ - {(1 + h)^2} - 1] = \mathop {\lim }\limits_{h \to 0} [ - 1 - {h^2} - 2h - 1]$
$= - {(0)^2} - 2 \times 0 - 2 = - 2$
$\therefore$ LHL $\neq$ RHL
Therefore limit of given function does not exist
View full question & answer→Question 512 Marks
Find $\mathop {\lim }\limits_{x \to 0} f(x)$ and $\mathop {\lim }\limits_{x \to 1} f(x)$ where $f(x) = \left\{ {\begin{array}{*{20}{c}} {2x + 3}&{x \le 0} \\ {3(x + 1)}&{x > 0} \end{array}} \right.$
AnswerHere f(x) = $\left\{ {\begin{array}{*{20}{c}} {2x + 3}&{x \le 0} \\ {3(x + 1)}&{x > 0} \end{array}} \right.$
Now $\mathop {\lim }\limits_{x \to 0} f(x)$ = $\mathop {\lim }\limits_{x \to 0} 2x + 3$ = 2 $\times$ 0 + 3 = 3
$\mathop {\lim }\limits_{x \to 1} f(x)$ = $\mathop {\lim }\limits_{x \to 1} 3(x + 1) $ = 3(1 + 1) = 3 $\times$ 2 = 6
View full question & answer→Question 522 Marks
Evaluate $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan 2x}}{{x - \frac{\pi }{2}}}$
AnswerHere $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan 2x}}{{x - \frac{\pi }{2}}}\left[ {\frac{0}{0}{\text{from}}} \right]$
Put $x = \frac{\pi }{2} + y$ as $x \to \frac{\pi }{2},\;y \to 0$
$\therefore \;\mathop {\lim }\limits_{y \to 0} \frac{{\tan 2\left( {\frac{\pi }{2} + y} \right)}}{{\frac{\pi }{2} + y - \frac{\pi }{2}}} = \mathop {\lim }\limits_{y \to 0} \frac{{\tan (\pi + 2y)}}{y}$
$ = \mathop {\lim }\limits_{y \to 0} \frac{{\tan 2y}}{y} = \mathop {\lim }\limits_{y \to 0} \frac{{\tan 2y}}{{2y}} \times 2 = 1 \times 2 = 2$
View full question & answer→Question 532 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} ({\text{cosec }}x - \cot x)$
AnswerHere $\mathop {\lim }\limits_{x \to 0} ({\text{cosec }}x - \cot x)$
$= \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{\sin x}} - \frac{{\cos x}}{{\sin x}}} \right)$
$= \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{\sin x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x/2}}{{2\sin x/2\cos x/2}} = \mathop {\lim }\limits_{x \to 0} \tan x/2 = 0$
View full question & answer→Question 542 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}};$ a, b, a + b $\neq$ 0.
AnswerWe have,
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin ax}}{x} + \frac{{bx}}{x}}}{{\frac{{ax}}{x} + \frac{{\sin bx}}{x}}}$
[dividing both numerator and denominator by x]
$ = \frac{{\mathop {\lim }\limits_{x \to 0} \frac{{\sin (ax)}}{{ax}} \times a + \mathop {\lim }\limits_{x \to 0} b}}{{\mathop {\lim }\limits_{x \to 0} a + \mathop {\lim }\limits_{x \to 0} \frac{{\sin bx}}{{bx}} \times b}}$$\left[ {\because \mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}{\text{ and }}} {\mathop {\lim }\limits_{x \to a} f(x) + g(x) = \mathop {\lim }\limits_{x \to a} f(x) + \mathop {\lim }\limits_{x \to a} g(x)} \right]$
$ = \frac{{a\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{ax}} + \mathop {\lim }\limits_{x \to 0} b}}{{\mathop {\lim }\limits_{x \to 0} a + b\mathop {\lim }\limits_{x \to 0} \frac{{\sin bx}}{{bx}}}}$
$= \frac { a ( 1 ) + b } { a + b ( 1 ) }$ $\left[ {\because \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1} \right]$
$= \frac { a + b } { a + b } = 1$
View full question & answer→Question 552 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} x\sec x$
AnswerHere $\mathop {\lim }\limits_{x \to 0} x\sec x$
$ = \mathop {\lim }\limits_{x \to 0} x \times \frac{1}{{\cos x}}\rightarrow\mathop {\lim }\limits_{x \to 0} \frac{x}{{\cos x}} = \frac{0}{1} = 0$
View full question & answer→Question 562 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}}.$
AnswerWe have, $\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}} = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{ax}}{{b\sin x}} + \frac{{x\cos x}}{{b\sin x}}} \right)$
$ = \frac{a}{b}\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sin x}} + \frac{1}{b}\mathop {\lim }\limits_{x \to 0} \frac{{x\cos x}}{{\sin x}}$
$ = \frac{a}{b}\mathop {\lim }\limits_{x \to 0} \frac{1}{{\left( {\frac{{\sin x}}{x}} \right)}} + \frac{1}{b}\mathop {\lim }\limits_{bx \to 0} \frac{{\cos x}}{{\left( {\frac{{\sin x}}{x}} \right)}}$
$ = \frac{a}{b}\frac{{\mathop {\lim }\limits_{x \to 0} 1}}{{\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right)}} + \frac{1}{b}\frac{{\mathop {\lim }\limits_{x \to 0} \cos x}}{{\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right)}}$$\left[\because {\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}} \right]$
$ = \frac{a}{b} \times \frac{1}{1} + \frac{1}{b} \times \frac{1}{1}$$\left[\because {\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1} \right]$
$= \frac { a + 1 } { b }$
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Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x - 1}}{{\cos x - 1}}$
AnswerHere $\mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x - 1}}{{\cos x - 1}}$$= \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 2x}}{{1 - \cos x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x}}{{2{{\sin }^2}x/2}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{(2\sin x/2\cos x/2)}^2}}}{{{{\sin }^2}x/2}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\sin }^2}x/2{{\cos }^2}x/2}}{{{{\sin }^2}x/2}} = \mathop {\lim }\limits_{x \to 0} 4{\cos ^2}x/2$ = 4/2 = 2
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Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{\pi - x}}$
AnswerHere $\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{\pi - x}} = \frac{{\cos 0}}{{\pi - 0}} = \frac{1}{\pi }$
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Evaluate $\mathop {\lim }\limits_{x \to \pi } \frac{{\sin (\pi - x)}}{{\pi (\pi - x)}}$
AnswerLet y= $\mathop {\lim }\limits_{x \to \pi } \frac{{\sin (\pi - x)}}{{\pi (\pi - x)}}\left[ {\frac{0}{0}{\text{from}}} \right]$
Put $x = \pi + y$, as $x \to \pi ,\;y \to 0$
$\therefore y=\;\mathop {\lim }\limits_{y \to 0} \frac{{\sin [\pi - \pi - y]}}{{\pi [\pi - \pi - y]}}=\mathop {\lim }\limits_{y \to 0} \frac{{\sin ( - y)}}{{ - \pi y}}$
$ = \mathop {\lim }\limits_{y \to 0} \frac{{ - \sin y}}{{ - \pi y}} = \frac{1}{\pi }\mathop {\lim }\limits_{y \to 0} \frac{{\sin y}}{y} = \frac{1}{\pi } \times1 = \frac{1}{\pi }$
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Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}},\;a,\;b \ne 0$
AnswerHere $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}}$$= \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin ax}}{{ax}} \times ax \times \frac{1}{{\frac{{\sin bx}}{{bx}} \times bx}}} \right]$
$ = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin ax}}{{ax}} \times \frac{1}{{\frac{{sinbx}}{{bx}}}} \times \frac{{ax}}{{bx}}} \right] = \frac{a}{b}$$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin ax}}{{ax}}\frac{1}{{\frac{{\sin bx}}{{bx}}}}} \right]$
$ = \frac{a}{b} \times 1 \times 1 = \frac{a}{b}$
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Evaluate $\lim \limits_{x \rightarrow 0} \frac{\sin a x}{b x}$
Answer Given, $\lim _{x \rightarrow 0} \frac{\sin a x}{b x}$
$$Applying the limits in the given expression we get,$\lim _{x \rightarrow 0} \frac{\sin a x}{b x}=\frac{0}{0}$
Multiplying and dividing the given expression by a we get,
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin a x}{b x} \times \frac{a}{a}$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin a x}{a x} \times \frac{a}{b}$
We know that: $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
$= \frac{a}{b} \lim _{a x \rightarrow 0} \frac{\sin a x}{a x}=\frac{a}{b} \times 1=\frac{a}{b}$
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Evaluate $\mathop {\lim }\limits_{x \to - 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}}$
AnswerHere $\mathop {\lim }\limits_{x \to - 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}}$
$ = \mathop {\lim }\limits_{x \to - 2} \frac{{\frac{{x + 2}}{{2x}}}}{{x + 2}}$
$ = \mathop {\lim }\limits_{x \to - 2} \frac{{x + 2}}{{2x}} \times \frac{1}{{x + 2}}$
$ = \mathop {\lim }\limits_{x \to - 2} \frac{1}{{2x}} = \frac{1}{{2 \times - 2}} = \frac{{ - 1}}{4}$
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Evaluate $\mathop {\lim }\limits_{x \to 1} \frac{{a{x^2} + bx + c}}{{c{x^2} + bx + a}},\;a + b + c \ne 0$
AnswerHere $\mathop {\lim }\limits_{x \to 1} \frac{{a{x^2} + bx + c}}{{c{x^2} + bx + a}}$
$ = \frac{{a \times {{(1)}^2} + b \times 1 + c}}{{c \times {{(1)}^2} + b \times 1 + a}} = \frac{{a + b + c}}{{c + b + a}} = 1$
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Evaluate $\mathop {\lim }\limits_{z \to 1} \frac{{{z^{1/3}} - 1}}{{{z^{1/6}} - 1}}$
AnswerHere $\mathop {\lim }\limits_{z \to 1} \frac{{{z^{1/3}} - 1}}{{{z^{1/6}} - 1}}\left[ {\frac{0}{0}{\text{form}}} \right]$
$= \mathop {\lim }\limits_{z \to 1} \frac{{{{({z^{1/6}} - 1)}^2}}}{{{z^{1/6}} - 1}}$
$ = \mathop {\lim }\limits_{z \to 1} \frac{{({z^{1/6}} + 1)({z^{1/6}} - 1)}}{{({z^{1/6}} - 1)}} = \mathop {\lim }\limits_{z \to 1}$
$= z^{1/6} + 1$
$= (1)^{1/6} + 1 = 1 + 1 = 2$
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Find the derivative of the following functions from first principle: $-\text{x}$
AnswerLet f(x) = –x. Accordingly, f(x + h) = -(x + h) By first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{-(\text{x+h})-(-\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{x}-\text{h}+\text{x}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}(-1)=-1$
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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): $4\sqrt{\text{x}}-2$
AnswerLet $\text{f(x)}=4\sqrt{\text{x}}-2$ $\text{f}'\text{(x)}=\frac{\text{d}}{\text{dx}}(4\sqrt{\text{x}}-2)=\frac{\text{d}}{\text{dx}}(4\sqrt{\text{x}})-\frac{\text{d}}{\text{dx}}(2)$ $=4\frac{\text{d}}{\text{dx}}\Big(\text{x}^{\frac{1}{2}}\Big)-0=4\Big(\frac{1}{2}\text{x}^{\frac{1}{2}-1}\Big)$ $=\Big(2\text{x}^{\frac{1}{2}}\Big)=\frac{2}{\sqrt{\text{x}}}$
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Find the derivative of. $2\text{x}-\frac{3}{4}$
AnswerLet $\text{f}(\text{x})=2\text{x}-\frac{3}{4}$ $=2\frac{\text{d}}{\text{dx}}(\text{x})-\frac{\text{d}}{\text{dx}}\bigg(\frac{3}{4}\bigg)$ $=2-0$ $=2$
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Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow1}\frac{\text{ax}^2+\text{bx}+\text{c}}{\text{cx}^2+\text{bx}+\text{a}},\text{a}+\text{b}+\text{c}\neq0$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\text{ax}^2+\text{bx}+\text{c}}{\text{cx}^2+\text{bx}+\text{a}}=\frac{\text{a}(1)^2+\text{b}(1)+\text{c}}{\text{c}(1)^2+\text{b}(1)+\text{a}}$$=\frac{\text{a}+\text{b}+\text{c}}{\text{a}+\text{b}+\text{c}}$
$=1 $ $[\text{a}+\text{b}+\text{c}\neq0]$
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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): $(\text{x}^2+1)\cos\text{x}$
AnswerLet $\text{f(x)}=(\text{x}^2+1)\cos\text{x}$ By product rule, $\text{f}'\text{(x)}=(\text{x}^2+1)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\cos\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^2+1)$ $=(\text{x}^2+1)(-\sin\text{x})+\cos\text{x}(2\text{x})$ $=-\text{x}^2\sin\text{x}-\sin\text{x}+2\text{x}\cos\text{x}$
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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): $\frac{1+\frac{1}{\text{x}}}{1-\frac{1}{\text{x}}}$
AnswerLet $\text{f(x)}=\frac{1+\frac{\text{1}}{\text{x}}}{1-\frac{1}{\text{x}}}=\frac{\frac{\text{x+1}}{\text{x}}}{\frac{\text{x}-1}{\text{x}}}=\frac{\text{x+1}}{\text{x}-1},\text{where x}\neq0$By quotient rule,
$\text{f}'\text{(x)}=\frac{(\text{x}-1)\frac{\text{d}}{\text{dx}}\text{(x+1)}-\text{(x+1)}\frac{\text{d}}{\text{dx}}(\text{x}-1)}{(\text{x}-1)^2},\text{x}\neq0, 1$
$=\frac{(\text{x}-1)(1)-\text{(x+1)}\text{(1)}}{(\text{x}-1)^2},\text{x}\neq0, 1$
$=\frac{\text{x}-1-\text{x}-1}{(\text{x}-1)^2},\text{x}\neq0, 1$
$=\frac{-2}{(\text{x}-1)^2},\text{x}\neq0,1$
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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): $\text{(x+a)}$
AnswerLet $\text{f(x)}=\text{x+a}$. Accordingly, $\text{f(x+h)}=\text{x+h+a}$ By first principle, $\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{x+h+a}-\text{x}-\text{a}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\text{h}}{\text{h}}\Big)$ $=\lim\limits_{\text{h}\rightarrow0}(1)$ $=1$
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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): $\frac{1}{\text{ax}^2+\text{bx+c}}$
AnswerLet $\text{f(x)}=\frac{1}{\text{ax}^2+\text{bx+c}}$ By quotient rule, $\text{f}'\text{(x)}=\frac{(\text{ax}^2\text{+bx+c})\frac{\text{d}}{\text{dx}}(1)-\frac{\text{d}}{\text{dx}}(\text{ax}^2\text{+bx+c})}{(\text{ax}^2\text{+bx+c})^2}$ $=\frac{(\text{ax}^2\text{+bx+c})(0)-(2\text{ax+b)}}{(\text{ax}^2\text{+bx+c})^2}$ $=\frac{-(2\text{ax+b})}{(\text{ax}^2\text{+bx+c})^2}$
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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): $\sin^{\text{n}}\text{x}$
AnswerLet $\text{y}=\sin^{\text{n}}\text{x}.$Accordingly, for n = 1, y = sin x.$\therefore\frac{\text{dy}}{\text{dx}}=\cos\text{x},\text{ i.e.,}\frac{\text{d}}{\text{dx}}\sin\text{x}=\cos\text{x}$
For n = 2, y = sin^2 x. $\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin\text{x}\sin\text{x})$
$=(\sin\text{x})'\sin\text{x}+\sin\text{x}(\sin\text{x})'\ [\text{By Leibnitz product rule]}$
$=\cos\text{x}\sin\text{x}+\sin\text{x}\cos\text{x}$
$=2\sin\text{x}\cos\text{x}\ ...(1)$For n = 3, y = sin^3 x.
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin\text{x}\sin^2\text{x})$
$=(\sin\text{x})'\sin^2\text{x}+\sin\text{x}(\sin^2\text{x})'\ [\text{By Leibnitz product rule]}$$=\cos\text{x}\sin^2\text{x}+\sin(2\sin\text{x}\cos\text{x})\ [\text{Using (1)]}$
$=\cos\text{x}\sin^2\text{x}+2\sin^2\text{x}\cos\text{x}$
$=3\sin^2\text{x}\cos\text{x}$ We assert that $\frac{\text{d}}{\text{dx}}(\sin^{\text{n}}\text{x})=\text{n}\sin^{(\text{n}-1)}\text{x}\cos\text{x}$Let our assertion be true for n = k. i. e., $\frac{\text{d}}{\text{dx}}(\sin^{\text{k}}\text{x})=\text{k}\sin^{(\text{k}-1)}\text{x}\cos\text{x}\ ...(2)$
$\text{Consider}\frac{\text{d}}{\text{dx}}(\sin^{\text{k}+1}\text{x})=\frac{\text{d}}{\text{dx}}(\sin\text{x}\sin^{\text{k}}\text{x})$
$=(\sin\text{x})'\sin^{\text{k}}\text{x}+\sin\text{x}(\sin^{\text{k}}\text{x})'\ [\text{By Leibnitz product rule]}$
$=\cos\text{x}\sin^{\text{k}}\text{x}+\sin\text{x}(\text{k}\sin^{(\text{k}-1)}\text{x}\cos\text{x})\ [\text{Using (2)]}$
$=\cos\text{x}\sin^{\text{k}}\text{x}+\text{k}\sin^{\text{k}}\text{x}\cos\text{x}$
$=(\text{k+1})\sin^{\text{k}}\text{x}\cos\text{x}$ Thus, our assertion is true for n = k + 1. Hence, by mathematical induction, $\frac{\text{d}}{\text{dx}}(\sin^{\text{n}}\text{x})=\text{n}\sin^{(\text{n}-1)}\text{x}\cos\text{x}$
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Find the derivative of . $\text{x}^5(3-6\text{x}^{-9})$
AnswerLet $\text{f}(\text{x})=\text{x}^5(3-6\text{x}^{-9})$ By Leibnitz product rule, $\text{f}'\big(\text{x})=\text{x}^5\frac{\text{d}}{\text{dx}}\big(3-6\text{x}^{-9}\big)+\big(3-6\text{x}^{-9}\big)\frac{\text{d}}{\text{dx}}\big(\text{x}^5\big)$ $=\text{x}^5\big\{0-6(-9)\text{x}^{-9-1}\big\}+(3-6\text{x}^{-9})(5\text{x}^4)$ $=\text{x}^5(54\text{x}^{-10})+15\text{x}^4-30\text{x}^{-5}$ $=54\text{x}^{-5}+15\text{x}^4-30\text{x}^{-5}$ $=24\text{x}^{-5}+15\text{x}^4$ $=15\text{x}^4+\frac{24}{\text{x}^5}$
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Find the derivative of. $(5\text{x}^3+3\text{x}-1)(\text{x}-1)$
AnswerLet $\text{f}(\text{x})=(5\text{x}^3+3\text{x}-1)(\text{x}-1)$ By Leibnitz product rule, $\text{f}'(\text{x})=(5\text{x}^3+3\text{x}-1)\frac{\text{d}}{\text{dx}}(\text{x}-1)+(\text{x}-1)\frac{\text{d}}{\text{dx}}(5\text{x}^3+3\text{x}-1)$ $=(5\text{x}^3+3\text{x}-1)(1)+(\text{x}-1)(5.3\text{x}^2+3\text{x}-0)$ $=(5\text{x}^3+3\text{x}-1)+(\text{x}-1)(15\text{x}^2+3)$ $=5\text{x}^3+3\text{x}-1+15\text{x}^3+3\text{x}-15\text{x}^2-3$ $=20\text{x}^3-15\text{x}^2+6\text{x}-4$
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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): $\big(\text{ax}^2+\sin\text{x}\big)(\text{p+q}\cos\text{x})$
AnswerLet $\text{f(x)}=\big(\text{ax}^2+\sin\text{x}\big)(\text{p+q}\cos\text{x})$ By product rule, $\text{f}'\text{(x)}=(\text{ax}^2+\sin\text{x})\frac{\text{d}}{\text{dx}}(\text{p+q}\cos\text{x})+(\text{p+q}\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{ax}^2\sin\text{x})$ $=(\text{ax}^2+\sin\text{x})(-\text{q}\sin\text{x})+(\text{p+q}\cos\text{x})(2\text{ax}+\cos\text{x})$ $=-\text{q}\sin\text{x}(\text{ax}^2+\sin\text{x})+(\text{p+q}\cos\text{x})(2\text{ax}+\cos\text{x})$
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Find the derivative of. $\text{x}^{-3}(5+3\text{x})$
AnswerLet $\text{f}(\text{x})=\text{x}^{-3}(5+3\text{x})$ By Leibnitz product rule, $\text{f}'\big(\text{x})=(\text{x}^{-3}\big)\frac{\text{d}}{\text{dx}}\big(5+3\text{x}\big)+\big(5+3\text{x}\big)\frac{\text{d}}{\text{dx}}\big(\text{x}^{-3}\big)$ $=\text{x}^{-3}(0+3)+(5+3\text{x})(-3\text{x}^{-3-1})$ $=\text{x}^{-3}(3)+(5+3\text{x})(-3\text{x}^{-4})$ $=3\text{x}^{-3}-15\text{x}^{-4}-9\text{x}^{-3}$ $=-6\text{x}^{-3}-15\text{x}^{-4}$ $=-3\text{x}^{-3}\Big(2+\frac{5}{\text{x}}\Big)$ $=\frac{-3\text{x}^{-3}}{\text{x}}\big(2{\text{x}+5}\big)$ $=\frac{-3}{\text{x}^4}\big(5+2{\text{x}}\big)$
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Find the derivative of . $\text{x}^{-4}(3-4\text{x}^{-5})$
AnswerLet $\text{f}(\text{x})=\text{x}^{-4}(3-4\text{x}^{-5})$ By Leibnitz product rule, $\text{f}'\big(\text{x})=\text{x}^{-4}\frac{\text{d}}{\text{dx}}\big(3-4\text{x}^{-5}\big)+\big(3-4\text{x}^{-5}\big)\frac{\text{d}}{\text{dx}}\big(\text{x}^{-4}\big)$ $=\text{x}^{-4}\big\{0-4(-5)\text{x}^{-5-1}\big\}+(3-4\text{x}^{-5})(-4)(\text{x}^{-4-1})$ $=\text{x}^{-4}(20\text{x}^{-6})+(3-4\text{x}^{-5})(-4\text{x}^{-5})$ $=20\text{x}^{-10}-12\text{x}^{-5}+16\text{x}^{-10}$ $=36\text{x}^{-10}-12\text{x}^{-5}$ $=-\frac{12}{\text{x}^5}+\frac{36}{\text{x}^{10}}$
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