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heureka

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 #2
avatar+26396 
+10

I. f(x)=arctan(8x);f(x)=?

\tan[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ ] = 8x \quad | \quad \frac{\ d()}{dx} \quad \small{\text{ and }} \quad \boxed{ [ \ tan(x)\ ]' = 1+\tan^2(x) }\\\\(1+\tan^2(\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ ) )\times\left[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ \right]' = 8 \\\\\left[ 1+(8x)^2} \right]\times\left[\textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ \right]' = 8 \\\\\boxed{ \left[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ \right]' = \frac{8} {1+(8x)^2} } }

II. f(x)=sin(x)cos(x);f(x)=?

y=sin(x)cos(x)|ln()ln(y)=cos(x)ln(sin(x))| d()dxyy=[cos(x)]ln(sin(x))+cos(x)[ln(sin(x))]y=y([cos(x)]ln(sin(x))+cos(x)[ln(sin(x))])y=sin(x)cos(x)(sin(x)ln(sin(x))+cos(x)cos(x)sin(x))y=sin(x)cos(x)(sin(x)ln(sin(x))+cos2(x)sin(x))

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13.01.2015
 #3
avatar+26396 
+10

 f(x)=(e+1)3x+1(x+2)sin2(3x+2) \samll f(x)=? $$ f(x)=(e+1)3x+1(x+2)sin2(3x+2)=(e+1)3(x+11x+21sin(3x+2)1sin(3x+2)) $$ f(x)=(e+1)3x+1(x+2)sin2(3x+2)((x+1)x+1(x+2)x+2(sin(3x+2))sin(3x+2)(sin(3x+2))sin(3x+2)) $$ f(x)=(e+1)3x+1(x+2)sin2(3x+2)(12x+1x+11x+23cos(3x+2)sin(3x+2)3cos(3x+2)sin(3x+2)) $$ f(x)=(e+1)3x+1(x+2)sin2(3x+2)(12x+1x+11x+223cos(3x+2)sin(3x+2)) $$ f(x)=(e+1)3x+1(x+2)sin2(3x+2)(12(x+1)1x+26cos(3x+2)sin(3x+2)) $$ P.S. (uv)=uv(uu+vv) and (uv)=uv(uuvv) and (uvw)=uvw(uuvvww) and (uvww)=uvww(uuvvwwww) 

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13.01.2015