1.कक्षा 11 में अवकलज (Derivatives in Class 11),अवकलज कक्षा 11 (Derivatives Class 11):

Derivatives in Class 11

कक्षा 11 में अवकलज (Derivatives in Class 11) के इस आर्टिकल में प्रथम सिद्धान्त से अवकलज तथा गुणन नियम जिसे Leibnitz नियम भी कहा जाता है,भागफल नियम द्वारा अवकलज ज्ञात करना सीखेंगे।
आपको यह जानकारी रोचक व ज्ञानवर्धक लगे तो अपने मित्रों के साथ इस गणित के आर्टिकल को शेयर करें।यदि आप इस वेबसाइट पर पहली बार आए हैं तो वेबसाइट को फॉलो करें और ईमेल सब्सक्रिप्शन को भी फॉलो करें।जिससे नए आर्टिकल का नोटिफिकेशन आपको मिल सके । यदि आर्टिकल पसन्द आए तो अपने मित्रों के साथ शेयर और लाईक करें जिससे वे भी लाभ उठाए । आपकी कोई समस्या हो या कोई सुझाव देना चाहते हैं तो कमेंट करके बताएं।इस आर्टिकल को पूरा पढ़ें।

Also Read This Article:-Limits Class 11

2.कक्षा 11 में अवकलज पर आधारित उदाहरण (Examples Based on Derivatives in Class 11):

Example:1.प्रथम सिद्धान्त से निम्नलिखित फलनों का अवकलज ज्ञात कीजिएः
Example:1(i).-x
Solution:माना f(x)=-x \\ f^{\prime}(x) =\underset{h \rightarrow 0}{\lim} \frac{f(x+h)-f(x)}{h} \\ =\underset{h \rightarrow 0}{\lim} \frac{\{-(x+h)\}-(-x)}{h} \\ =\underset{h \rightarrow 0}{\lim} \frac{-x-h+x}{h} \\ =\underset{h \rightarrow 0}{\lim} \left(\frac{-h}{h}\right)=\underset{h \rightarrow 0}{\lim}(-1) \\ \Rightarrow f^{\prime}(x) =-1
Example:1(ii). (-x)^{-1}
Solution:माना f(x)=(-x)^{-1} \\ f^{\prime}(x)=\underset{h \rightarrow 0}{\lim}\frac{f(x+h)-f(x)}{h} \\ =\underset{h \rightarrow 0}{\lim} \frac{\frac{1}{-(x+h)}-\left(-\frac{1}{x}\right)}{h} \\ =\underset{h \rightarrow 0}{\lim} \frac{-\frac{1}{x+h}+\frac{1}{x}}{h} \\ =\underset{h \rightarrow 0}{\lim} \frac{-x+x+h}{x(x+h)h} \\ =\underset{h \rightarrow 0}{\lim} \frac{h}{x(x+h) h} \\ =\underset{h \rightarrow 0}{\lim} \frac{1}{x(x+h)} \\ \Rightarrow f^{\prime}(x)=\frac{1}{x^2}
Example:1(iii). \sin (x+1)
Solution:माना f(x)=\sin (x+1) \\ f^{\prime}(x) =\underset{h \rightarrow 0}{\lim} \frac{f(x+h)-f(x)}{h} \\ =\underset{h \rightarrow 0}{\lim} \frac{\sin (x+h+1)-\sin (x+1)}{h} \\ =\underset{h \rightarrow 0}{\lim} \frac{\sin (x+1) \cos h+\cos (x+1) \sinh -\sin (x+1)}{h} \\ =\underset{h \rightarrow 0}{\lim} \sin (x+1) \frac{\cos h}{h}+\cos (x+1) \underset{h \rightarrow 0}{\lim} \left(\frac{\sin h}{h}\right) -\underset{h \rightarrow 0}{\lim} \frac{\sin (x+1)}{h} \\ =-\sin (x+1) \underset{h \rightarrow 0}{\lim} \left[\frac{1-\cos h}{h}\right]+\cos (x+1)-1 \\ =-\sin (x+1)(0)+\cos (x+1) \\ \Rightarrow f^{\prime}(x)=\cos (x+1)
Example:1(iv). \cos \left(x-\frac{\pi}{8}\right)
Solution: \cos \left(x-\frac{\pi}{8}\right)

माना f(x)=\cos \left(x-\frac{\pi}{8}\right) \\ f^{\prime}(x) =\underset{h \rightarrow 0}{\lim} \frac{f(x+h)-f(x)}{h} \\ =\underset{h \rightarrow 0}{\lim} \frac{\cos (x+h-\frac{\pi}{8})-\cos \left(x-\frac{\pi}{8}\right)}{h} \\ \underset{h \rightarrow 0}{\lim} \frac{\cos \left(x-\frac{\pi}{8}\right) \cos h-\sin \left(x-\frac{\pi}{8}\right) \sin h-\cos \left(x-\frac{\pi}{8}\right)}{h} \\ =\lim _{h \rightarrow 0} \left\{-\cos\left(x-\frac{\pi}{8}\right) \right\} \left(\frac{1-\cos h}{h}\right)-\sin \left(x-\frac{\pi}{8}\right) \underset{h \rightarrow 0}{\lim} \left(\frac{\sin h}{h}\right ) \\ =-\cos \left(x-\frac{\pi}{8}\right)(0)-\sin \left(x-\frac{\pi}{8}\right) \cdot 1 \\ \Rightarrow f^{\prime}(x)=-\sin \left(x-\frac{\pi}{8}\right)
निम्नलिखित फलनों के अवकलज ज्ञात कीजिए (यह समझा जाय कि a,b,c,d,p,q,r और s निश्चित शून्येतर अचर है और m तथा n पूर्णांक हैं।):
Example:2.(x+a)
Solution:माना y=(x+a) \\ \frac{d y}{d x}=\frac{d}{d x}(x)+\frac{d}{d x}(a) \\ \Rightarrow \frac{d y}{d x}=1+0=1
Example:3. (p x+q)\left(\frac{r}{x}+s\right)
Solution:माना y=(p x+q)\left(\frac{r}{x}+s\right) \\ \frac{d y}{d x} =\left(\frac{r}{x}+s\right) \frac{d}{d x}(p x+q)+(p x+q) \frac{d}{dx}\left(\frac{r}{x}+s\right) \\ =\left(\frac{r}{x}+s\right)(p \cdot 1)+(p x+q)\left(-\frac{r}{x^2}\right) \\ =\frac{p r}{x}+p s-\frac{p r}{x}-\frac{r q}{x^2} \\ \Rightarrow \frac{d y}{d x} =-\frac{q r}{x^2}+p s
Example:4. (a x+b)(c x+d)^2
Solution:माना y=(a x+b)(c x+d)^2 \\ \frac{d y}{d x}=(a x+b) \frac{d}{d x}(cx+d)^2+(c x+d)^2 \frac{d}{d x}(a x+b) \\ =(a x+b) \cdot 2(c x+d) \frac{d}{d x}(cx+d)+(cx+d)^2(a \cdot 1) \\ =2(a x+b)(c x+d)(c .1)+(c x+b)^2 \cdot a \\ \Rightarrow \frac{d y}{d x}=2 c(a x+b)(c x+d)+a(c x+b)^2
Example:5. \frac{a x+b}{c x+b}
Solution:माना y=\frac{a x+b}{c x+b} \\ \frac{d y}{d x} =\frac{(c x+b)(a x+b)^{\prime}-(a x+b)(c x+ b)^{\prime}}{(cx+b)^2}\left[\because \left(\frac{u}{v}\right)^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^{2}}\right] \\ =\frac{(c x+b) a-(a x+b) c}{(c x+b)^2} \\ =\frac{a c x+a b-acx-b c}{(c x+b)^2} \\ \Rightarrow \frac{d y}{d x} =\frac{b(a-c)}{(c x+b)^2}
Example:6. \frac{1+\frac{1}{x}}{1-\frac{1}{x}}
Solution:माना y=\frac{1+\frac{1}{x}}{1-\frac{1}{x}} \\ \Rightarrow y =\frac{1+x}{x-1} \\ \frac{d y}{d x} =\frac{(x-1)(1+x)^{\prime}-(1+x)(x-1)^{\prime}}{(x-1)^2}\left[\left(\frac{u}{v}\right)^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^{2}}\right] \\ =\frac{(x-1) \cdot 1-(1+x) \cdot 1}{(x-1)^2} \\ =\frac{x-1-1-x}{(x-1)^2} \\ \Rightarrow \frac{d y}{d x} =-\frac{2}{(x-1)^2}
Example:7. \frac{1}{a x^2+b x+c}
Solution:माना y=\frac{1}{a x^2+b x+c} \\ \frac{d y}{d x}=\frac{d}{d x}\left(a x^2+b x+c\right)^{-1} \\ =-1\left(a x^2+b x+c\right)^{-2} \frac{d}{d x}\left(a x^2+b x+c\right) \\ =-\left(a x^2+b x+c\right)^{-2}(2 a x+b) \\ \Rightarrow \frac{d y}{d x}=\frac{-(2 a x+b)}{\left(a x^2+b x+c\right)^2}
Example:8. \frac{a x+b}{p x^2+q x+r}
Solution:माना y=\frac{a x+b}{p x^2+q x+r} \\ \frac{d y}{d x}=\frac{\left(p x^2+q x+r\right)(a x+ b)^{\prime} -(a x+b)\left(p x^2+q x+r\right)^{\prime}}{\left(p x^2+q x+r\right)^2} \\ =\frac{\left(p x^2+q x+r\right) a-(a x+b)(2 p x+q)}{\left(p x^2+q x+r\right)^2} \\=\frac{\left(a p x^2+a q x+a r-2 a p x^2-aq x-2 b p x-b q\right)}{\left(p x^2+q x+r\right)^2} \\ \Rightarrow \frac{d y}{d x}= \frac{-a p x^2-2 b p x+a r-b q}{\left(p x^2+q x+r\right)^2}
Example:9. y=\frac{b x^2+q x+r}{a x+b}
Solution:माना y=\frac{p x^2+qx+r}{ax+b} \\ \frac{d y}{d x} =\frac{(a x+b)\left(b x^2+q x+r \right)^{\prime} -\left(p x^2+q x+r\right)(a x+b)^{\prime}}{(a x+b)^2} \\ =\frac{(a x+b)(2 p x+q)-\left(p x^2+q x+r\right) a}{(a x+b)^2}\\ =\frac{2 a b x^2+a q x+2 b p x+b q-a p x^2-a q x-a x}{(a x+b)^2} \\ \Rightarrow \frac{d y}{d x}= \frac{a p x^2+2 b p x+b q-a r}{(a x+b)^2}
Example:10. \frac{a}{x^4}-\frac{b}{x^2}+\cos x
Solution:माना y=\frac{a}{x^4}-\frac{b}{x^2}+\cos x \\ \frac{d y}{d x} =a \frac{d}{d x}\left(x^{-4}\right)-\frac{d}{d x} b \left(x^{-2}\right)+\frac{d}{d x}(\cos x) \\ =-4 x^{-5}+2 b x^{-3}-\sin x \\ \Rightarrow \frac{d y}{d x} =-\frac{4}{x^5}+\frac{2 b}{x^3}-\sin x
Example:11. 4 \sqrt{x}-2
Solution:माना y=4 x^{\frac{1}{2}}-2 \\ \frac{d y}{d x} =4 \frac{d}{d x}\left(x^{\frac{1}{2}}\right)-\frac{d}{d x} (2) \\ =4 \times \frac{1}{2} x^{-\frac{1}{2}}-(0) \\ \Rightarrow \frac{d y}{d x} =\frac{2}{\sqrt{x}}

Derivatives in Class 11

Example:12. (a x+b)^n
Solution:माना y=(a x+b)^n \\ \frac{d y}{d x} =\frac{d}{d x}(a x+b)^n \\=n(a x+b)^{n-1} \frac{d}{d x}(a x+b) \\ =n(a x+b)^{n-1}(a) \\ \Rightarrow \frac{d y}{d x} =a n(a x+b)^{n-1}
Example:13. (a x+b)^n(c x+d)^m
Solution:माना y=(a x+b)^n(c x+d)^m \\ \frac{d y}{d x}=(a x+b)^n \frac{d}{d x}(c x+d)^m+(c x+d)^m \frac{d}{d x}(a x+b)^n \\ =(a x+b)^n m(c x+d)^{m-1} \frac{d}{d x}(c x+d)+(c x+d)^{m} n(a x+b)^{n-1} \frac{d}{d x}(a x+b) \\ =(a x+b)^n(c x+d)^{m-1} \cdot c+(cx+d)^m \cdot n(a x+b)^{n-1} \cdot a \cdot 1 \\ =(a x+b)^{n-1} \left(c x+d\right)^{m-1}\left[c m(a x+b)+an(c x+d)\right] \\ \Rightarrow \frac{d y}{d x}=(a x+b)^{n-1}(cx+d)^{m-1}\left[acx(m+n)+ cmb+and\right]
Example:14. \sin (x+a)
Solution:माना y=\sin (x+a) \\ \frac{d y}{d x} =\frac{d}{d x} \sin (x+a) \\ =\cos (x+a) \frac{d}{d x}(x+a) \\=\cos (x+a) \cdot 1 \\ \Rightarrow \frac{d y}{d x} =\cos (x+a)
Example:15. \operatorname{cosec} x \cot x
Solution:माना y=\operatorname{cosec} x \cot x \\ \frac{d y}{d x} =\operatorname{cosec} x \frac{d}{d x}(\cot x)+\cot x \frac{d}{dx}(\operatorname{cosec} x) \\=\operatorname{cosec} x \cdot \left( \operatorname{cosec}^2 x\right)+\cot x(-\operatorname{cose} x \cot x) \\ \Rightarrow \frac{d y}{d x} =-\operatorname{cosec} x\left(\operatorname{cosec}^2 x+\cot ^2 x\right)
Example:16. \frac{\cos x}{1+\sin x}
Solution:माना y =\frac{\cos x}{1+\sin x} \\ \frac{d y}{d x} =\frac{(1+\sin x)(\cos x)^{\prime}-\cos x \frac{d}{d x}(1+\sin x)}{(1+\sin x)^2} \\ =\frac{(1+\sin x)(-\sin x)-\cos x \cdot \cos x}{(1+\sin x)^2} \\ =\frac{-\sin x-\left(\sin ^2 x+\cos ^2 x\right)}{(1+\sin x)^2} \\ \frac{d y}{d x}=-\frac{(1+\sin x)}{(1+\sin x)^2} \\ \Rightarrow \frac{d y}{d x}=-\frac{1}{1+\sin x}
Example:17. \frac{\sin x+\cos x}{\sin x-\cos x}
Solution:माना y=\frac{\sin x+\cos x}{\sin x-\cos x} \\ \frac{d y}{d x}=\frac{(\sin x-\cos x)(\sin x+\cos x)^{\prime}-(\sin x+\cos x)(\sin x-\cos x)^{\prime}}{(\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-\sin ^2 x-\cos ^2 x+\sin x \cos x-\sin ^2 x-\cos ^2 x-2 \sin x \cos x}{(\sin x-\cos x)^2} \\ =\frac{-2\left(\sin ^2 x+\cos ^2 x\right)}{(\sin x-\cos x)^2} \\ \Rightarrow \frac{d y}{d x}=\frac{-2}{(\sin x-\cos x)^2}
Example:18. \frac{\sec x-1}{\sec x+1}
Solution:माना y=\frac{\sec x-1}{\sec x+1} \\ \frac{d y}{d x}=\frac{(\sec x+1)(\sec x-1)^{\prime}-(\sec x-1)(\sec x+1)^{\prime}}{(\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} \\ \Rightarrow \frac{d y}{d x}=\frac{2 \sec x \tan x}{(\sec x+1)^2}
Example:19. \sin ^n x
Solution:माना y=\sin ^n x \\ \frac{d y}{d x}=n \sin ^{n-1} x \frac{d}{d x}(\sin x) \\ \Rightarrow \frac{d y}{d x}=n \sin ^{n-1} x \cos x
Example:20. \frac{a+b \sin x}{c+d \cos x}
Solution:माना y=\frac{a+b \sin x}{c+d \cos x} \\ \frac{d y}{d x}=\frac{(c+d \cos x)(a+b \sin x)^{\prime}-(a+b \sin x)(c+d \cos x)^{\prime}}{(c+d \cos x)^{\prime}} \\ =\frac{(c+d \cos x)(b \cos x)-(a+b \sin x) (-d \sin x)}{(c+d \cos x)^2} \\=\frac{b c \cos x+b d \cos ^2 x+a d \sin x+b d \sin ^2 x}{ (c+d \cos x)^2} \\ = \frac{b c \cos x+a d \sin x+b d\left(\cos ^2 x+\sin ^2 x\right)}{(c+d \cos x)^2} \\ \Rightarrow \frac{d y}{d x}= \frac{b c \cos x+a d \sin x+b d}{(c+d \cos x)^2}
Example:21. \frac{\sin (x+a)}{\cos x}
Solution:माना y=\frac{\sin (x+a)}{\cos x} \\ \frac{d y}{d x}=\frac{\cos x[\sin (x+a)]^{\prime}-\sin (x+a)(\cos x)^{\prime}}{\cos ^2 x} \\ = \frac{\cos x \cos (x+a)+\sin (x+a) \sin x}{\cos ^2 x} \\ =\frac{\cos (x+a-x)}{\cos ^2 x} \\ \Rightarrow \frac{d y}{d x} =\frac{\cos a}{\cos ^2 x}
Example:22. x^4(5 \sin x-3 \cos x)
Solution:माना y=x^4(5 \sin x-3 \cos x) \\ \frac{d y}{d x} =x^4 \frac{d}{d x}(5 \sin x-3 \cos x)+(5 \sin x-3 \cos x)\frac{d}{d x} x^4 \\ =x^4(5 \cos x+3 \sin x)+(5 \sin x-3 \cos x) \cdot 4 x^3 \\ \Rightarrow \frac{d y}{d x} =x^3(5 x-12) \cos x+x^3(3 x+20) \sin x
Example:23. \left(x^2+1\right) \cos x
Solution:माना y=\left(x^2+1\right) \cos x \\ \frac{d y}{d x} =\left(x^2+1\right) \frac{d}{d x} \cos x+\cos x \frac{d}{d x}\left(x^2+1\right) \\ =\left(x^2+1\right)(-\sin x)+\cos x \cdot 2 x \\ \Rightarrow \frac{d y}{d x} =-\left(x^2+1\right) \sin x+2 x \cos x
Example:24. \left(a x^2+\sin x\right)(p+q \cos x)
Solution:माना y=\left(a x^2+\sin x\right)(p+q \cos x) \\ \frac{d y}{d x}=\left(a x^2+\sin x\right) \frac{d}{d x}(p+q \cos x)+(p+q \cos x) \frac{d}{d x}\left(a x^2+\sin x\right) \\ =\left(a x^2+\sin x\right)(-q \sin x)+(p+q \cos x)(2 a x+\cos x) \\ = -a q x^2 \sin x-q \sin ^2 x+2 a p x+p \cos x +2 a q x \cos x+2 a q x \cos^{2} x \\ \Rightarrow \frac{d y}{d x}= -a q x^2 \sin x-q \sin ^2 x+2 a q x \cos ^2 x +2 a q x \cos x+p \cos x+2 a p x
Example:25. (x+\cos x)(x-\tan x)
Solution:माना y=(x+\cos x)(x-\tan x) \\ \frac{d y}{d x}= (x+\cos x) \frac{d}{d x}(x-\tan x)+(x-\tan x) \frac{d}{d x}(x+\cos x) \\ = (x+\cos x)\left(1-\sec ^2 x\right)+(x-\tan x)(1-\sin x) \\ =x-x \sec ^2 x+\cos x-\sec x+x-x \sin x-\tan x+\sin x \tan x \\ \Rightarrow \frac{d y}{d x}= (x+\cos x)-\sec ^2 x(x+\cos x)+x (1-\sin x)-\tan (1-\sin x) \\ \Rightarrow \frac{d y}{d x}= (x+\cos x)\left(1-\sec ^2 x\right)+(1-\sin x) (x-\tan x)
Example:26. \frac{4 x+5 \sin x}{3 x+7 \cos x}
Solution:माना y=\frac{4 x+5 \sin x}{3 x+7 \cos x} \\ \frac{d y}{d x}=\frac{(3 x+7 \cos x)(4 x+5 \sin x)^{\prime}-(4 x+5 \sin x)(3 x+7 \cos x)^{\prime}}{(3 x+7 \cos x)^2} \\ =\frac{(3 x+7 \cos x)(4+5 \cos x)-(4 x+5 \sin x)(3-7 \sin x)}{(3 x+7 \cos x)^2}\\ =\frac{(15 x+28) \cos x+(28 x-15) \sin x +35\left(\cos ^2 x+\sin ^2 x\right)}{(3 x+7 \cos x)^2} \\ \Rightarrow \frac{d y}{d x}=\frac{(15 x+28) \cos x+(28 x+5) \sin x+35}{(3 x+7 \cos x)^2}
Example:27. \frac{x^2 \cos \left(\frac{\pi}{4}\right)}{\sin x}
Solution:माना y=\frac{x^2 \cos \left(\frac{\pi}{4}\right)}{\sin x} \\ \Rightarrow y =\frac{x \cdot \frac{1}{\sqrt{2}}}{\sin x} \\ \frac{d y}{d x} =\frac{1}{\sqrt{2}} \frac{\sin x \frac{d}{d x}(x)-x \frac{d}{d x}(\sin x)}{\sin ^2 x} \\ =\frac{1}{\sqrt{2}} \frac{\sin x \cdot 1-x \cos x}{\sin ^2 x} \\ \Rightarrow \frac{d y}{d x}=\frac{\sin x-x \cos x}{\sqrt{2} \sin ^2 x}
Example:28. \frac{x}{1+\tan x}
Solution:माना y=\frac{x}{1+\tan x} \\ \frac{d y}{d x} =\frac{(1+\tan x) \frac{d}{d x}(x)-x \frac{d}{d}(1+\tan x)}{(1+\tan x)^2} \\ =\frac{(1+\tan x) \cdot 1-x \sec ^2 x}{(1+\tan x)^2} \\ \Rightarrow \frac{d y}{d x} =\frac{1+\tan x-x \sec ^2 x}{(1+\tan x)^2}
Example:29. (x+\sec x)(x-\tan x)
Solution:माना y=(x+\sec x)(x-\tan x) \\ \frac{d y}{d x}=(x+\sec x) \frac{d}{d x}(x-\tan x)+(x-\tan x)\frac{d}{d x}(x+\sec x) \\ \Rightarrow \frac{d y}{d x}= (x+\sec x)\left(1-\sec ^2 x\right)+(x-\tan x) (1+\sec x \tan x)
Example:30. \frac{x}{\sin ^n x}
Solution:माना y=\frac{x}{\sin ^n x} \\ \frac{d y}{d x}=\frac{\sin ^n x \frac{d}{d x}(x)-x \frac{d}{d x}\left(\sin ^n x\right)}{\sin ^{2 n} x} \\ =\frac{\sin ^n x \cdot 1-x \cdot n \sin ^{n-1} x \cos x}{\sin ^{2 n} x} \\ \Rightarrow \frac{d y}{d x}=\frac{\sin ^{n-1} x\left(\sin ^x x-n x \cos x\right)}{\sin ^{2 n} x}
उपर्युक्त उदाहरणों के द्वारा कक्षा 11 में अवकलज (Derivatives in Class 11),अवकलज कक्षा 11 (Derivatives Class 11) को समझ सकते हैं।

3.कक्षा 11 में अवकलज पर आधारित सवाल (Questions Based on Derivatives in Class 11):

Derivatives in Class 11

निम्नलिखित फलनों के अवकलज ज्ञात कीजिएः

(1.) \sqrt{\frac{1-\cos x}{1+\cos x}}
(2.) x^2 e^{x} \sin x
(3.) \operatorname{cosec} x+2^{x+3}+4 \log _3 x
उत्तर (Answers): (1.) \operatorname{cosec} x(\operatorname{cosec} x-\cot x)
(2.) x e^x(2 \sin x+x \sin x+x \cos x)
(3.) -\operatorname{cosec} x \cot x+2^{x+3} \log _e 2+\frac{4}{x} \log _3 e
उपर्युक्त सवालों को हल करने पर कक्षा 11 में अवकलज (Derivatives in Class 11),अवकलज कक्षा 11 (Derivatives Class 11) को ठीक से समझ सकते हैं।

Also Read This Article:-Derivatives Class 11

4.कक्षा 11 में अवकलज (Frequently Asked Questions Related to Derivatives in Class 11),अवकलज कक्षा 11 (Derivatives Class 11) से सम्बन्धित अक्सर पूछे जाने वाले प्रश्न:

कक्षा 11 में अवकलज (Derivatives in Class 11) के इस आर्टिकल में प्रथम सिद्धान्त से
अवकलज तथा गुणन नियम जिसे Leibnitz नियम भी कहा जाता है