Home › STD11-12 SCIENCE STD11-12 SCIENCE વિકલન April 17, 2017 Leave a Reply Posted By: Aakash Kavaiya A+ A- સામાન્ય નિયમ{\displaystyle \left({cf}\right)'=cf'}{\displaystyle \left({f+g}\right)'=f'+g'}ગુણાકારનો નિયમ{\displaystyle \left({fg}\right)'=f'g+fg'}ભાગાકારનો નિયમ{\displaystyle \left({f \over g}\right)'={f'g-fg' \over g^{2}},\qquad g\neq 0}શૃખંલા નિયમ{\displaystyle (f\circ g)'=(f'\circ g)g'}કેટલાક સામાન્ય વિકલન{\displaystyle {d \over dx}c=0}{\displaystyle {d \over dx}x=1}{\displaystyle {d \over dx}(cx)=c}{\displaystyle {d \over dx}|x|={|x| \over x}=\operatorname {sgn} x,\qquad x\neq 0}{\displaystyle {d \over dx}x^{c}=cx^{c-1}\qquad {\mbox{where both }}x^{c}{\mbox{ and }}cx^{c-1}{\mbox{ are defined}}}{\displaystyle {d \over dx}\left({1 \over x}\right)={d \over dx}\left(x^{-1}\right)=-x^{-2}=-{1 \over x^{2}}}{\displaystyle {d \over dx}\left({1 \over x^{c}}\right)={d \over dx}\left(x^{-c}\right)=-cx^{-c-1}=-{c \over x^{c+1}}}{\displaystyle {d \over dx}{\sqrt {x}}={d \over dx}x^{1 \over 2}={1 \over 2}x^{-{1 \over 2}}={1 \over 2{\sqrt {x}}},\qquad x>0}ચરઘાતાંકિય અને લઘુગણકીય વિકલન{\displaystyle {d \over dx}c^{x}={c^{x}\ln c},\qquad c>0}{\displaystyle {d \over dx}e^{x}=e^{x}}{\displaystyle {d \over dx}\log _{c}x={1 \over x\ln c},\qquad c>0,c\neq 1}{\displaystyle {d \over dx}\ln x={1 \over x},\qquad x>0}{\displaystyle {d \over dx}\ln |x|={1 \over x}}{\displaystyle {d \over dx}x^{x}=x^{x}(1+\ln x)}ત્રિકોણમિતિય વિકલન{\displaystyle {d \over dx}\sin x=\cos x}{\displaystyle {d \over dx}\cos x=-\sin x}{\displaystyle {d \over dx}\tan x=\sec ^{2}x={1 \over \cos ^{2}x}}{\displaystyle {d \over dx}\sec x=\tan x\sec x}{\displaystyle {d \over dx}\csc x=-\csc x\cot x}{\displaystyle {d \over dx}\cot x=-\csc ^{2}x={-1 \over \sin ^{2}x}}{\displaystyle {d \over dx}\arcsin x={1 \over {\sqrt {1-x^{2}}}}}{\displaystyle {d \over dx}\arccos x={-1 \over {\sqrt {1-x^{2}}}}}{\displaystyle {d \over dx}\arctan x={1 \over 1+x^{2}}}{\displaystyle {d \over dx}\operatorname {arcsec} x={1 \over |x|{\sqrt {x^{2}-1}}}}{\displaystyle {d \over dx}\operatorname {arccsc} x={-1 \over |x|{\sqrt {x^{2}-1}}}}{\displaystyle {d \over dx}\operatorname {arccot} x={-1 \over 1+x^{2}}}હાઇપરબોલિક વિકલન{\displaystyle {d \over dx}\sinh x=\cosh x={\frac {e^{x}+e^{-x}}{2}}}{\displaystyle {d \over dx}\cosh x=\sinh x={\frac {e^{x}-e^{-x}}{2}}}{\displaystyle {d \over dx}\tanh x=\operatorname {sech} ^{2}\,x}{\displaystyle {d \over dx}\,\operatorname {sech} \,x=-\tanh x\,\operatorname {sech} \,x}{\displaystyle {d \over dx}\,\operatorname {coth} \,x=-\,\operatorname {csch} ^{2}\,x}{\displaystyle {d \over dx}\,\operatorname {csch} \,x=-\,\operatorname {coth} \,x\,\operatorname {csch} \,x}{\displaystyle {d \over dx}\,\operatorname {arcsinh} \,x={1 \over {\sqrt {x^{2}+1}}}}{\displaystyle {d \over dx}\,\operatorname {arccosh} \,x={1 \over {\sqrt {x^{2}-1}}}}{\displaystyle {d \over dx}\,\operatorname {arctanh} \,x={1 \over 1-x^{2}}}{\displaystyle {d \over dx}\,\operatorname {arcsech} \,x={-1 \over x{\sqrt {1-x^{2}}}}}{\displaystyle {d \over dx}\,\operatorname {arccoth} \,x={1 \over 1-x^{2}}}{\displaystyle {d \over dx}\,\operatorname {arccsch} \,x={-1 \over |x|{\sqrt {1+x^{2}}}}}પ્રતિવિધેયનુ વિકલન{\displaystyle {d \over dx}(f^{-1}(x))={\frac {1}{f'(f^{-1}(x))}}} Share On Facebook Share On twitter
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