{"id":4182,"date":"2024-01-30T12:23:57","date_gmt":"2024-01-30T06:23:57","guid":{"rendered":"https:\/\/stage-wp.10minuteschool.com\/?p=4182"},"modified":"2024-11-05T15:39:07","modified_gmt":"2024-11-05T09:39:07","slug":"limit-for-different-functions","status":"publish","type":"post","link":"https:\/\/10minuteschool.com\/content\/limit-for-different-functions\/","title":{"rendered":"\u09ab\u09be\u0982\u09b6\u09a8\u09c7\u09b0 \u099c\u09a8\u09cd\u09af \u09b2\u09bf\u09ae\u09bf\u099f \u098f\u09b0 \u09ae\u09be\u09a8 \u09a8\u09bf\u09b0\u09cd\u09a3\u09af\u09bc (Determining the value of the limit for different functions)"},"content":{"rendered":"

Type 04- \u09a4\u09cd\u09b0\u09bf\u0995\u09cb\u09a3\u09ae\u09bf\u09a4\u09bf \u09ab\u09be\u0982\u09b6\u09a8\u09c7\u09b0<\/b>\u00a0\u09b2\u09bf\u09ae\u09bf\u099f<\/b><\/a> (Limit of trigonometry function):<\/b><\/span><\/h3>\n

Q. \\frac{\\lim }{x \\rightarrow 0} \\frac{1-\\cos 7 x}{3 x^{2}}<\/span><\/p>\n\n\n\\begin{array}{l}\n\n=\\frac{\\lim }{x \\rightarrow 0} \\frac{2 \\sin ^{2} \\frac{7 x}{2}}{3 x^{2}} \\\\\n\n=\\frac{2}{3} \\frac{\\lim }{x \\rightarrow 0}\\left(\\frac{\\sin \\frac{7 x}{2}}{\\frac{7 x}{2}}\\right)^{2} \\times\\left(\\frac{7 x}{2}\\right)^{2} \\times \\frac{1}{x^{2}} \\\\\n\n=\\frac{2}{3} \\times \\frac{7^{2}}{4}=\\frac{49}{6}\n\n\\end{array}\n\n<\/span>\n

(Ans)<\/strong><\/p>\n

Q. \\frac{\\lim }{x \\rightarrow 0} \\frac{\\tan x-\\sin x}{x^{3}}<\/span><\/p>\n\n\n\\begin{array}{l}\n\n=\\frac{\\lim }{x \\rightarrow 0} \\frac{\\tan x(1-\\cos x)}{x^{3}} \\\\\n\n=\\frac{\\lim }{x \\rightarrow 0} \\frac{\\tan x \\times 2 \\sin ^{2} \\frac{x}{2}}{x^{3}} \\\\\n\n=\\frac{\\lim }{x \\rightarrow 0} \\frac{\\tan x}{x} \\\\\n\n=\\frac{\\lim }{x \\rightarrow 0} \\frac{\\sin x}{x} \\frac{x}{2} \\\\\n\n=2 \\times \\frac{1}{4}=\\frac{1}{2} \\\\\n\n\\end{array}\n\n<\/span>\n

(Ans)<\/strong><\/p>\n

Example-27: \\frac{\\lim }{x \\rightarrow 0} \\frac{\\cos 7 x-\\cos 9 x}{\\cos 3 x-\\cos 5 x}<\/span><\/strong><\/p>\n

Sol ^{n}: \\frac{\\lim }{x \\rightarrow 0} \\frac{\\cos 7 x-\\cos 9 x}{\\cos 3 x-\\cos 5 x}<\/span><\/p>\n\n\n\\begin{array}{l}\n\n=\\frac{\\lim }{x \\rightarrow 0} \\frac{\\cos 7 x-\\cos 9 x}{\\cos 3 x-\\cos 5 x} \\\\\n\n=\\frac{\\frac{\\lim }{x \\rightarrow 0} \\frac{\\sin 8 x}{8 x} \\cdot 8 x}{\\frac{\\lim \\sin 4 x}{x \\rightarrow 0} 4 x} \\cdot 4 x \\\\\n\n=2\n\n\\end{array}\n\n<\/span>\n

 <\/p>\n

(Ans)<\/strong><\/p>\n

Type-05: x<\/span> \u098f\u09b0 \u09ae\u09be\u09a8 \u0985\u09b8\u09c0\u09ae\u09c7\u09b0 \u09a6\u09bf\u0995\u09c7 \u09a7\u09be\u09ac\u09bf\u09a4 \u09b9\u09b2\u09c7 \u09b2\u09bf\u09ae\u09bf\u099f\u09c7\u09b0<\/a> \u09ae\u09be\u09a8 \u09a8\u09bf\u09b0\u09cd\u09a3\u09df <\/b>(Determining the value of limit if the value of x<\/span> runs towards infinity)<\/b> :<\/b><\/span><\/h3>\n

Example-13: \\frac{L t}{x \\rightarrow \\infty} \\frac{a x^{2}+b x+c}{L x^{2}+m x+n}<\/span><\/b>\u098f\u09b0 \u09ae\u09be\u09a8 \u0995\u09a4?<\/b><\/p>\n

\\mathrm{Sol}^{n}<\/span>:<\/span> \\frac{L t}{x \\rightarrow \\infty} \\frac{\\left(a+\\frac{b}{x}+\\frac{c}{x^{2}}\\right)}{\\left(l+\\frac{m}{x}+\\frac{n}{x^{2}}\\right)} <\/span>=\\frac{a+0+0}{l+0+0} <\/span>= x <\/span><\/p>\n

(Ans)<\/strong><\/p>\n

Example-14: \\frac{\\lim }{x \\rightarrow \\infty} \\frac{1^{2}+2^{2}+3^{2}+----+x^{2}}{x^{3}+x^{2}+x+1} <\/span><\/b>\u00a0\u098f\u09b0 \u09ae\u09be\u09a8 \u0995\u09a4?<\/b><\/p>\n

\\mathrm{Sol}^{n}<\/span>: \\frac{\\lim }{x \\rightarrow \\infty} \\frac{1^{2}+2^{2}+3^{2}+----+x^{2}}{x^{3}+x^{2}+x+1} <\/span><\/span><\/p>\n

= \\frac{\\lim }{x \\rightarrow \\infty} \\frac{x(x+1)(2 x+1)}{6\\left(x^{3}+x^{2}+x+1\\right)} <\/span><\/span><\/p>\n

<\/i>= \\frac{\\lim }{x \\rightarrow \\infty} \\frac{2 x^{3}+2 x^{2}+x}{6 x^{3}+6 x^{2}+6 x+6}<\/span><\/span><\/p>\n

= \\frac{\\lim }{x \\rightarrow \\infty} \\frac{2+\\frac{3}{x}+\\frac{1}{x^{2}}}{6+\\frac{6}{x}+\\frac{6}{x^{2}}+6 \/ x^{3}} <\/span><\/span><\/p>\n

= \\frac{2+0+0}{6+0+0+0} <\/span><\/span>= \\frac{2}{6} <\/span>= \\frac{1}{3} <\/span><\/p>\n

(Ans)<\/strong><\/p>\n

Q. \\frac{\\lim }{x \\rightarrow \\infty} \\frac{3^{x}-3^{-x}}{3^{x}+3^{-x}}<\/span><\/p>\n

\\begin{array}{l}\n\n=\\frac{\\lim }{x \\rightarrow \\infty} \\frac{3^{x}\\left(1-3^{-2 x}\\right)}{3^{x}\\left(1+3^{-2 x}\\right)} \\\\\n\n=\\frac{\\lim }{x \\rightarrow \\infty} \\frac{1-\\frac{1}{3^{2 x}}}{1-\\frac{1}{3^{2 x}}} \\\\\n\n=1 \\\\\n\n\\end{array}<\/span>\u00a0<\/span><\/p>\n

(Ans)<\/strong><\/p>\n\\begin{array}{l}\n\n\\text { Q. } \\frac{\\lim }{n \\rightarrow \\infty} \\frac{1}{n^{4}} \\sum_{r=1}^{n} \\boldsymbol{r}^{3} \\\\\n\n=\\frac{\\lim }{n \\rightarrow \\infty} \\frac{1}{n^{4}}\\left\\{1+2^{3}+3^{3}+----n^{3}\\right\\} \\\\\n\n=\\frac{\\lim }{n \\rightarrow \\infty} \\frac{1}{n^{4}}\\left\\{\\frac{n(n+1)}{2}\\right\\}^{2} \\\\\n\n=\\frac{\\lim }{n \\rightarrow \\infty} \\frac{1}{n^{4}}\\left\\{\\frac{n^{2}\\left(1+\\frac{1}{n}\\right)^{2}}{4}\\right\\} \\\\\n\n=\\frac{1}{4} \\cdot \\frac{\\lim }{n \\rightarrow \\infty} \\frac{1}{n^{2}}\\left(1+\\frac{1}{n}\\right)^{2} \\\\\n\n=\\frac{1}{4} \\cdot 1=\\frac{1}{4} \\\\\n\n\\end{array}<\/span>\n

(Ans)<\/strong><\/p>\n

Type-05: Inverse circular function <\/b>\u098f\u09b0<\/b> \u09b2\u09bf\u09ae\u09bf\u099f<\/b> \u09a8\u09bf\u09b0\u09cd\u09a3\u09df<\/b> (Determine the limit of inverse circular function) :<\/b><\/span><\/h3>\n

Concept: (i) Inverse circular functiom <\/span>\u09a5\u09be\u0995\u09ac\u09c7<\/span> ; (ii) <\/span>\u09ac\u09c0\u099c\u0997\u09be\u09a3\u09bf\u09a4\u09bf\u0995<\/span> \u09ac\u09be<\/span> \u09a4\u09cd\u09b0\u09bf\u0995\u09cb\u09a3\u09cd\u09ae\u09bf\u09a4\u09bf\u0995<\/span> \u09ab\u09be\u0982\u09b6\u09a8<\/span> ;\u00a0<\/span>(iii) <\/span>lim<\/span>x\u21920<\/span> \u09a5\u09be\u0995\u09ac\u09c7<\/span>\u0964<\/span><\/p>\n

Example- 39. <\/b>\u09ae\u09be\u09a8 \u09a8\u09bf\u09b0\u09cd\u09a3\u09df \u0995\u09b0\u0983 \\frac{L t}{x \\rightarrow 0} \\frac{\\tan ^{-1} x}{x}<\/span><\/b><\/p>\n

Sol ^{n}<\/span>:\u00a0<\/span> \u09a7\u09b0\u09bf, \\tan ^{-1} x=\\theta<\/span><\/span> \u09ac\u09be \\tan \\theta=x<\/span> \u09af\u09c7\u09b9\u09c7\u09a4\u09c1 x \\rightarrow 0<\/span> \u09a4\u09be\u0987 \\theta \\rightarrow 0<\/span><\/p>\n\\therefore \\frac{L t}{x \\rightarrow 0} \\frac{\\tan ^{-1} x}{x}=\\frac{L t}{\\theta \\rightarrow 0} \\frac{\\theta}{\\tan \\theta}=\\frac{L t}{\\theta \\rightarrow 0} \\frac{\\frac{1}{\\tan \\theta}}{\\theta}=1<\/span>\n

(Ans)\u00a0<\/strong><\/p>\n

\u27f9 \u09ab\u09be\u0982\u09b6\u09a8\u09c7\u09b0 \u0985\u09a8\u09cd\u09a4\u09b0\u09c0\u0995\u09b0\u09a3 \u09af\u09cb\u0997\u09cd\u09af\u09a4\u09be\u0983<\/strong>\u00a0<\/span><\/span><\/h3>\n

\u098f\u0995\u099f\u09bf \u09ab\u09be\u0982\u09b6\u09a8 \u09a4\u0996\u09a8\u09bf \u0985\u09a8\u09cd\u09a4\u09b0\u09c0\u0995\u09b0\u09a3<\/span> \u09af\u09cb\u0997\u09cd\u09af<\/span> \u09b9\u09ac\u09c7<\/span> \u09af\u0996\u09a8<\/span> \u09ab\u09be\u0982\u09b6\u09a8\u099f\u09bf<\/span> \u0985\u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8<\/span> \u09b9\u09ac\u09c7<\/span>\u0964<\/span> \u09ac\u09bf\u099a\u09cd\u099b\u09bf\u09a8\u09cd\u09a8<\/span> \u09ab\u09be\u0982\u09b6\u09a8\u09c7\u09b0<\/span> \u0985\u09a8\u09cd\u09a4\u09b0\u09c0\u0995\u09b0\u09a3<\/span>\u00a0 <\/span>\u09b8\u09ae\u09cd\u09ad\u09ac<\/span> \u09a8\u09df<\/span>\u0964<\/span><\/p>\n

\u27f9 \u09aa\u09cd\u09b0\u09be\u09df\u09cb\u0997\u09bf\u0995 \u09ad\u09be\u09b7\u09be\u09df \u0985\u09a8\u09cd\u09a4\u09b0\u09c0\u0995\u09b0\u09a3 \u098f\u0995\u099f\u09bf\u00a0 \u09b0\u09c8\u0996\u09bf\u0995 \u09aa\u09cd\u09b0\u0995\u09cd\u09b0\u09bf\u09df\u09be \u09af\u09be \u0987\u09a8\u09aa\u09c1\u099f \u0995\u09b0\u09c7 \u098f\u0995\u099f\u09bf \u09ab\u09be\u0982\u09b6\u09a8 \u098f\u09ac\u0982 \u0986\u0989\u099f\u09aa\u09c1\u099f \u0995\u09b0\u09c7 \u0985\u09aa\u09b0 \u098f\u0995\u099f\u09bf \u09ab\u09be\u0982\u09b6\u09a8 \u098f\u09b0\u09c2\u09aa \u09af\u09c7\u09a8, \u09aa\u09cd\u09b0\u09a4\u09bf\u099f\u09bf \u09ac\u09bf\u09a8\u09cd\u09a6\u09c1\u09a4\u09c7 \u0986\u0989\u099f\u09aa\u09c1\u099f\u09c7\u09b0 \u09ae\u09be\u09a8 \u09b9\u09df \u0987\u09a8\u09aa\u09c1\u099f \u09a2\u09be\u09b2\u0964<\/p>\n

\"\u09ab\u09be\u0982\u09b6\u09a8\"<\/p>\n

\u2234 <\/span>\u09a2\u09be\u09b2<\/span> = \\frac{\\Delta y}{\\Delta x} <\/span><\/span>= \\frac{f(x+h)-f(x)}{x+h-x}<\/span><\/span>=\\frac{f(x+h)-f(x)}{h}<\/span><\/span><\/p>\n

\u09af\u0996\u09a8 x<\/span> \u098f\u09b0 \u09aa\u09b0\u09bf\u09ac\u09b0\u09cd\u09a4\u09a8 \u0985\u09a4\u09bf\u0995\u09cd\u09b7\u09c1\u09a6\u09cd\u09b0 \u0985\u09b0\u09cd\u09a5\u09be\u09ce <\/span>h\u21920 <\/span> <\/span>\u09a4\u0996\u09a8<\/span>\u00a0 x <\/span><\/span>\u098f\u09b0<\/span> \u09b8\u09be\u09aa\u09c7\u0995\u09cd\u09b7\u09c7<\/span>\u00a0 y<\/span> <\/span>\u098f\u09b0<\/span> \u09af\u09c7\u00a0\u09aa\u09b0\u09bf\u09ac\u09b0\u09cd\u09a4\u09a8<\/span> \u09b9\u09df<\/span> \u09a4\u09be\u0995\u09c7<\/span> y<\/span> <\/span>\u098f\u09b0<\/span> \u0985\u09a8\u09cd\u09a4\u09b0\u099c<\/span> \u09ac\u09b2\u09c7<\/span>\u0964<\/span><\/p>\n

\u0985\u09b0\u09cd\u09a5\u09be\u09ce,\u00a0<\/span><\/p>\n

\\frac{\\lim }{h \\rightarrow 0} \\frac{f(x+h)-f(x)}{h}<\/span> <\/span>\u27f6 <\/span>\u09ae\u09c2\u09b2<\/span> \u09a8\u09bf\u09df\u09ae\u09c7<\/span> \u0985\u09a8\u09cd\u09a4\u09b0\u099c<\/span> \u09b8\u09c2\u09a4\u09cd\u09b0<\/span>\u0964<\/span><\/p>\n

\\frac{\\lim }{h \\rightarrow 0} \\frac{f(x+h)-f(x)}{h}<\/span>=\\frac{d y}{d x}<\/span><\/span>\u00a0<\/span><\/p>\n

\\frac{d y}{d x} <\/span>=<\/span> \u09a2\u09be\u09b2 <\/span>=m <\/span>=\\tan \\theta <\/span><\/span>=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}} <\/span><\/span>= \\frac{\\Delta y}{\\Delta x}<\/span><\/span>= f^{\\prime}(x)=y_{1} <\/span><\/span><\/p>\n

\u09ae\u09c2\u09b2\u09a8\u09bf\u09df\u09ae\u09c7\u09b0 \u0985\u09a8\u09cd\u09a4\u09b0\u09c0\u0995\u09b0\u09a3\u0983<\/strong><\/span><\/h3>\n

(i)<\/b> \\frac{d}{d x}^{(\\sin x)}<\/span>\u00a0\u09a8\u09bf\u09b0\u09cd\u09a3\u09df\u0983<\/strong><\/p>\n\\begin{array}{l}\n\n\\frac{d}{d x}^{(\\sin x)}=\\frac{\\lim }{h \\rightarrow 0} \\frac{\\sin (x+h)-\\sin x}{h} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times 2 \\cos \\frac{x+h+x}{2} \\cdot \\sin \\frac{x+h-x}{2} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times 2 \\cos \\frac{2 x+h}{2} \\cdot \\sin \\frac{h}{2} \\\\\n\n=2 \\times \\frac{\\lim }{\\frac{h}{2} \\rightarrow 0} \\frac{\\sin \\frac{h}{2}}{\\frac{h}{2}} \\times \\frac{h}{2} \\times \\frac{\\lim }{h \\rightarrow 0} \\cos \\frac{2 x+h}{2} \\times \\frac{1}{h} \\\\\n\n=2 \\times \\frac{\\lim }{h \\rightarrow 0} \\frac{h}{2} \\times \\frac{1}{h} \\times \\cos \\frac{2 x+h}{2} \\\\\n\n=\\cos \\frac{2 x}{2}=\\cos x \\\\\n\n\\therefore \\frac{d}{d x}(\\sin x)=\\cos x\n\n\\end{array}<\/span>\n

 <\/p>\n

(ii) \\frac{d}{d x}^{(\\cos x)}<\/span><\/b>\u00a0<\/strong>\u00a0 \u09a8\u09bf\u09b0\u09cd\u09a3\u09df\u0983\u00a0<\/b><\/p>\n\\begin{array}{l}\n\n\\frac{d}{d x}^{(\\cos x)}=\\frac{\\lim }{h \\rightarrow 0} \\frac{\\cos (x+h)-\\cos x}{h} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times 2 \\sin \\frac{x+h+x}{2} \\cdot \\sin \\frac{x-h-x}{2} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times 2 \\sin \\frac{2 x+h}{2} \\cdot \\sin \\frac{-h}{2} \\\\\n\n=-2 \\times \\frac{\\lim }{\\frac{h}{2} \\rightarrow 0} \\frac{\\sin \\frac{h}{2}}{\\frac{h}{2}} \\times \\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times \\frac{h}{2} \\sin \\frac{2 x+h}{2} \\\\\n\n=-2 \\times \\frac{1}{2} \\sin \\frac{2 x}{2}=-\\sin x \\\\\n\n\\therefore \\frac{d}{d x}^{(\\cos x)}=-\\sin x\n\n\\end{array}<\/span>\n

 <\/p>\n

(iii) \\frac{d}{d x}^{(\\tan x)} <\/span><\/b>\u00a0 \u00a0\u09a8\u09bf\u09b0\u09cd\u09a3\u09df\u0983\u00a0<\/strong><\/p>\n

\\begin{array}{l}\n\n\\frac{d}{d x}^{(\\tan x)}=\\frac{\\lim }{h \\rightarrow 0} \\frac{\\tan (x+h)-\\tan x}{h} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times\\left(\\frac{\\sin (x+h)}{\\cos (x+h)}-\\frac{\\sin x}{\\cos x}\\right) \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times \\frac{\\sin (x+h) \\cdot \\cos x-\\sin x \\cdot \\cos (x+h)}{\\cos (x+h) \\cdot \\cos x} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h \\cos (x+h) \\cdot \\cos x} \\times \\sin (x+h-x) \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{\\sin h}{h} \\times \\frac{1}{\\cos (x+h) \\cdot \\cos x} \\\\\n\n=\\frac{1}{\\cos ^{2} x}=\\sec ^{2} x \\\\\n\n\\therefore \\frac{d}{d x}(\\tan x)\n\n\\end{array}<\/span>\u00a0 <\/span>\u00a0<\/span><\/p>\n

 <\/p>\n

(iv) \\frac{d}{d x}^{(\\cot x)}<\/span><\/b>\u00a0 \u09a8\u09bf\u09b0\u09cd\u09a3\u09df\u0983\u00a0<\/strong><\/p>\n

\\begin{array}{l}\n\n\\frac{d}{d x}^{(\\cot x)} \\quad=\\frac{\\lim }{h \\rightarrow 0} \\frac{\\cot (x+h)-\\cot x}{h} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times\\left(\\frac{\\cos (x+h)}{\\sin (x+h)}-\\frac{\\cos x}{\\sin x}\\right) \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times\\left(\\frac{\\cos (x+h) \\cdot \\sin x-\\cos x \\cdot \\sin (x+h)}{\\sin (x+h) \\cdot \\sin x}\\right) \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times \\frac{\\sin (x-x-h)}{\\sin (x+h) \\cdot \\sin x} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\times \\frac{-\\sin h}{h \\sin (x+h) \\cdot \\sin x} \\\\\n\n=-\\frac{\\lim }{h \\rightarrow 0} \\frac{\\sin h}{h} \\times \\frac{\\lim }{h \\rightarrow 0} \\frac{h}{\\operatorname{hsin}(x+h) \\cdot \\sin x} \\\\\n\n=-\\frac{1}{\\sin ^{2} x}=-\\operatorname{cosec}^{2} x \\\\\n\n\\therefore \\frac{d}{d x}(\\cot x) \\\\\n\n\\end{array}<\/span> <\/span>\u00a0<\/span><\/p>\n

 <\/p>\n

(v) \\frac{d}{d x}^{(\\sec x)} <\/span><\/b>\u00a0 \u09a8\u09bf\u09b0\u09cd\u09a3\u09df\u0983\u00a0<\/strong><\/p>\n

\\begin{array}{l}\n\n\\frac{d}{d x}^{(\\sec x)}=\\frac{\\lim }{h \\rightarrow 0} \\frac{\\sec (x+h)-\\sec x}{h} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times\\left(\\frac{1}{\\cos (x+h)}-\\frac{1}{\\cos x}\\right) \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times \\frac{\\cos x-\\cos (x+h)}{\\cos (x+h) \\cdot \\cos x} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{\\cos (x+h) \\cdot \\cos x} \\cdot 2 \\sin \\frac{2 x+h}{2} \\cdot \\frac{\\lim }{h \\rightarrow 0} \\cdot \\sin \\frac{x+h-x}{2} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{\\sin \\frac{h}{2}}{\\frac{h}{2}} \\times \\frac{\\lim }{h \\rightarrow 0} \\frac{2 \\sin \\left(x+\\frac{h}{2}\\right)}{\\cos (x+h) \\cdot \\cos x} \\cdot \\times \\frac{2}{2} \\\\\n\n=\\frac{1}{\\cos ^{2} x} \\cdot \\sin \\frac{2 x}{2} \\\\\n\n=\\frac{\\sin x}{\\cos x} \\times \\frac{1}{\\cos x} \\\\\n\n=\\tan x \\cdot \\sec x \\\\\n\n\\therefore \\frac{d}{d x}(\\sec x)\n\n\\end{array}<\/span>\u00a0 <\/span>\u00a0\u00a0<\/span><\/p>\n

 <\/p>\n

(vi) \\frac{d}{d x}^{(\\operatorname{cosec} x)}<\/span><\/b>\u00a0\u09a8\u09bf\u09b0\u09cd\u09a3\u09df\u0983\u00a0<\/b><\/p>\n

\\begin{array}{l}\n\n\\frac{d}{d x}^{(\\operatorname{cosec} x)}=\\frac{\\lim }{h \\rightarrow 0} \\frac{\\operatorname{cosec}(x+h)-\\operatorname{cosec} x}{h} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times\\left(\\frac{1}{\\sin (x+h)}-\\frac{1}{\\sin x}\\right) \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times \\frac{\\sin x-\\sin (x+h)}{\\sin (x+h) \\sin x} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\times \\frac{1}{h \\sin (x+h) \\sin x} \\times 2 \\cos \\frac{x+x+h}{2} \\cdot \\sin \\frac{x-x-h}{2} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\times \\frac{1}{h \\sin (x+h) \\sin x} \\times 2 \\cos \\frac{2 x+h}{2}-\\sin \\frac{h}{2} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{\\sin \\frac{h}{2}}{h} \\times \\frac{\\lim }{h \\rightarrow 0} \\frac{h}{2} \\times \\frac{2 \\cos \\frac{2 x+h}{2}}{h \\sin (h+x) \\cdot \\sin x} \\\\\n\n=-\\frac{\\cos \\frac{2 x}{2}}{\\sin ^{2} x}=-\\frac{\\cos x}{\\sin x} \\times \\frac{1}{\\sin x} \\\\\n\n=-\\cot x . \\operatorname{cosec} x \\\\\n\n\\therefore \\frac{(\\operatorname{cosec} x)}{d x}\n\n\\end{array}<\/span> <\/span>\u00a0<\/span><\/p>\n

 <\/p>\n

(vii) \\frac{d}{d x}^{e^{x}}<\/span><\/b>\u00a0\u09a8\u09bf\u09b0\u09cd\u09a3\u09df\u0983\u00a0<\/strong><\/p>\n

\\begin{array}{l}\n\n\\frac{d e^{x}}{d x}=\\frac{\\lim }{h \\rightarrow 0} \\frac{e^{x+h}-e^{x}}{h} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{e^{x} \\cdot e^{h}-e^{x}}{h} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times e^{x}\\left(e^{h}-1\\right) \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{e^{x}}{h} \\times\\left(1+h+\\frac{h^{2}}{2 !}+\\frac{h^{3}}{3 !}+-----1\\right) \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{e^{x}}{h} \\times h\\left(1+\\frac{h^{2}}{2 !}+\\frac{h^{3}}{3 !}+--\\right) \\\\\n\n=\\mathrm{e}^{x} \\\\\n\n\\therefore \\frac{d}{d x}^{x^{x}}=e^{x}\n\n\\end{array}<\/span>\u00a0<\/span><\/p>\n

 <\/p>\n

(viii) \\frac{d}{d x}^{x^{n}}<\/span><\/b>\u00a0\u09a8\u09bf\u09b0\u09cd\u09a3\u09df\u0983\u00a0<\/strong><\/p>\n\\begin{array}{l}\n\n\\frac{d x^{n}}{d x}=\\frac{\\lim }{h \\rightarrow 0} \\frac{(x+h)^{n}-x^{n}}{h} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times\\left[\\left\\{x\\left(1+\\frac{h}{x}\\right)\\right\\}^{n}\\right] \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times\\left\\{x^{n}\\left(1-\\frac{h}{x}\\right)^{n}-x^{n}\\right\\} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times\\left\\{x^{n}\\left(1-\\frac{h}{x}\\right)^{n}-x^{n}\\right\\} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times x^{n}\\left\\{\\left(1-\\frac{h}{x}\\right)^{n}-1\\right\\} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{x^{n}}{h}\\left\\{1+\\frac{n h}{x}+\\frac{n(n-1)}{2 !} \\frac{h^{2}}{x^{2}}+\\frac{n(n-1)(n-2)}{3 !} \\frac{h^{3}}{x^{3}}+-----1\\right\\} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{x^{n}}{h} \\times \\frac{n h}{x}\\left\\{1+\\frac{(n-1)}{2 !} \\frac{h}{x}+\\frac{(n-1)(n-2)}{3 !} \\frac{h^{2}}{x^{2}}+----\\right\\} \\\\\n\n=n x^{n} \\times \\frac{1}{x}=n x^{n-1} \\\\\n\n\\therefore \\frac{d}{d x}^{x^{n}}=n x^{n-1}\n\n\\end{array}<\/span>\n

 <\/p>\n

(ix) \\frac{d}{d x}^{a^{x}}<\/span> <\/b>\u09a8\u09bf\u09b0\u09cd\u09a3\u09df\u0983\u00a0<\/b><\/p>\n\\begin{array}{l}\n\n\\frac{d^{a^{x}}}{d x}=\\frac{\\lim }{\\mathrm{h} \\rightarrow 0} \\frac{\\mathrm{a}^{\\mathrm{x}+h}-\\mathrm{a}}{h} \\\\\n\n=\\frac{\\lim }{\\mathrm{h} \\rightarrow 0} \\frac{\\mathrm{a}^{\\mathrm{x}} \\cdot \\mathrm{a}^{\\mathrm{h}}-\\mathrm{a}^{\\mathrm{x}}}{h} \\\\\n\n=\\frac{\\lim }{\\mathrm{h} \\rightarrow 0} \\frac{\\mathrm{a}^{\\mathrm{x}}\\left(\\mathrm{a}^{\\mathrm{h}}-1\\right)}{h} \\\\\n\n=\\frac{\\lim }{\\mathrm{h} \\rightarrow 0} \\frac{\\mathrm{a}^{\\mathrm{x}}}{h}\\left\\{1+h \\ln a+\\frac{h^{2}}{2 !}(\\ln a)^{2}+\\frac{h^{3}}{3 !}(\\ln a)^{3}+----\\quad-1\\right\\} \\\\\n\n=\\frac{\\lim }{\\mathrm{h} \\rightarrow 0} \\frac{\\mathrm{a}^{\\mathrm{x}}}{h} \\times h \\ln a\\left\\{1+\\frac{h^{2}}{2 !}(\\ln a)+\\frac{h^{3}}{3 !}(\\ln a)^{2}+----\\right\\} \\\\\n\n=a^{x} \\ln a \\\\\n\n\\therefore \\frac{d}{d x} a^{x}=a^{x} \\ln a\n\n\\end{array}<\/span>\n

 <\/p>\n

(x)<\/b> \\frac{d}{d x}^{\\ln x}<\/span><\/strong><\/span>\u00a0\u09a8\u09bf\u09b0\u09cd\u09a3\u09df\u0983\u00a0<\/strong><\/p>\n

\\begin{array}{l}\n\n\\frac{d}{d x} \\ln x \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{\\ln \\left(\\frac{x+h}{x}\\right)}{h} \\\\\n\n=\\frac{\\lim (x+h)-\\ln x}{h \\rightarrow 0} \\frac{1}{h} \\times \\ln \\left(1+\\frac{h}{x}\\right) \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times\\left\\{\\frac{h}{x}-\\frac{1}{2} \\frac{h^{2}}{x^{2}}+\\frac{1}{3} \\frac{h^{3}}{x^{3}}-\\frac{1}{4} \\frac{h^{4}}{x^{4}}+----\\right\\} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times\\left\\{\\frac{h}{x}-\\frac{1}{2} \\frac{h^{2}}{x^{2}}+\\frac{1}{3} \\frac{h^{3}}{x^{3}}-\\frac{1}{4} \\frac{h^{4}}{x^{4}}+----\\right\\} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times \\frac{h}{x}\\left\\{1-\\frac{1}{2} \\frac{h}{x}+\\frac{1}{3} \\frac{h^{2}}{x^{2}}-\\frac{1}{4} \\frac{h^{3}}{x^{3}}+----\\right\\} \\\\\n\n=\\frac{1}{x} \\\\\n\n\\therefore \\frac{d}{d x} \\ln x\n\n\\end{array}<\/span>\u00a0<\/span><\/p>\n

 <\/p>\n

(xi) \\frac{d}{d x} \\log _{a} x<\/span><\/b>\u00a0\u09a8\u09bf\u09b0\u09cd\u09a3\u09df\u0983\u00a0<\/strong><\/p>\n

\\begin{array}{l}\n\n\\frac{d}{d x} \\log _{a} x=\\frac{\\lim }{h \\rightarrow 0} \\frac{\\log _{a}(x+h)-\\log _{a} x}{h} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{\\log _{a}\\left(\\frac{x+h}{x}\\right)}{h} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{\\log _{a}\\left(1+\\frac{h}{x}\\right)}{h} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times \\log _{a} c \\times \\log _{c}\\left(1+\\frac{h}{x}\\right) \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h} \\times \\frac{1}{\\log _{c} a} \\times \\ln _{c}\\left(1+\\frac{h}{x}\\right) \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h \\ln a} \\times\\left\\{\\frac{h}{x}-\\frac{1}{2} \\frac{h^{2}}{x^{2}}+\\frac{1}{3} \\frac{h^{3}}{x^{3}}-----------\\right\\} \\\\\n\n=\\frac{\\lim }{h \\rightarrow 0} \\frac{1}{h \\ln a} \\times \\frac{h}{x}\\left\\{1-\\frac{1}{2} \\frac{h}{x}+\\frac{1}{3} \\frac{h^{2}}{x^{2}}-\\right. \\\\\n\n=\\frac{1}{x \\ln a}=\\frac{1}{x} \\times \\frac{1}{\\log _{c} a}=\\frac{1}{x} \\times \\log _{a} c \\\\\n\n\\therefore \\frac{d}{d x} \\log _{a} x \\\\\n\n=\\frac{1}{x \\ln a}=\\frac{1}{x} \\times \\log a c\n\n\\end{array}<\/span>\u00a0<\/span><\/p>\n

\u09b8\u09c2\u09a4\u09cd\u09b0<\/b> (Formula)<\/b> :<\/b><\/h3>\n

1. \\frac{d}{d x}^{\\{c f(x)\\}}=c \\frac{d}{d x}^{f(x)}<\/span> ; c = \u09a7\u09cd\u09b0\u09c1\u09ac\u0995<\/p>\n

2. \\frac{d^{c}}{d x}=0 ; \\quad \\mathrm{c}=<\/span> \u09a7\u09cd\u09b0\u09c1\u09ac\u0995<\/p>\n

3. \\frac{d}{d x}^{(u+v)}=\\frac{d}{d x}^{u}+\\frac{d^{v}}{d x}<\/span><\/p>\n

4. \\frac{d}{d x}^{(u-v)}=\\frac{d}{d x}^{u}-\\frac{d^{v}}{d x}<\/span><\/p>\n

5. \\frac{d}{d x}^{(u v)}=u \\frac{d}{d x}^{v}+v \\frac{d}{d x}^{u}<\/span><\/p>\n

6. \\frac{d}{d x}\\left(\\frac{u}{v}\\right)=\\frac{v \\frac{d}{d x}^{u}-u \\frac{d}{d x}^{v}}{v^{2}}<\/span><\/p>\n

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\u098f\u0987\u099a\u098f\u09b8\u09b8\u09bf \u0993 \u098f\u09a1\u09ae\u09bf\u09b6\u09a8 \u09aa\u09b0\u09c0\u0995\u09cd\u09b7\u09be\u09b0\u09cd\u09a5\u09c0\u09a6\u09c7\u09b0 \u099c\u09a8\u09cd\u09af \u0986\u09ae\u09be\u09a6\u09c7\u09b0 \u0995\u09cb\u09b0\u09cd\u09b8\u09b8\u09ae\u09c2\u09b9\u0983<\/strong><\/em><\/p>\n<\/div>\n<\/div>\n<\/div>\n