Đạo hàm và phương trình y'=0

<span class='text_page_counter'>(1)</span>5/> y:= (1/3*(cos(x)^4)(sin(x))^4); 1 y := cos( x ) 4  sin( x ) 4 3. 1/> y:= 2*sin(x)+cos(2*x); y := 2 sin( x )  cos( 2 x ). > y1:=diff(y,x); 4 y1 :=  cos( x ) 3 sin( x )  4 sin( x ) 3 cos( x ) 3. > y1:=diff(y,x); y1 := 2 cos( x )  2 sin( 2 x ) > y2:=solve(y1=0,{x}); 1 1 1 5 y2 := { x    }, { x   }, { x   }, { x   } 2 2 6 6. > y2:=solve(y1=0,{x}); 1 1 { x  0 }, { x    }, { x   } 2 2. 2/> y:= (1/3*(sin(x)^3)-(cos(x))^2); 1 y := sin( x ) 3  cos( x ) 2 3. 6/ > y:= (cos(x)^4)-(sin(x))^4; y := cos( x ) 4  sin( x ) 4 > y1:=diff(y,x); y1 := 4 cos( x ) 3 sin( x )  4 sin( x ) 3 cos( x ). > y1:=diff(y,x); y1 := sin( x ) 2 cos( x )  2 cos( x ) sin( x ). > y2:=solve(y1=0,{x}); 1 1 y2 := { x  0 }, { x    }, { x   } 2 2. > y2:=solve(y1=0,{x}); 1 1 { x  0 } { x    }, { x   } 2 2. 7/ > y:= (cos(x)^6)+(sin(x))^6; y := cos( x ) 6  sin( x ) 6. 3/ > y:=. (1/3*(cos(x)^3)+(sin(x))^2); 1 y := cos( x ) 3  sin( x ) 2 3. > y1:=diff(y,x); y1 := 6 cos( x ) 5 sin( x )  6 sin( x ) 5 cos( x ). > y1:=diff(y,x); y1 := cos( x ) 2 sin( x )  2 cos( x ) sin( x ). > y2:=solve(y1=0,{x}); 1 1 { x  0 }, { x    }, { x   } 2 2. > y2:=solve(y1=0,{x}); 1 1 { x  0 }, { x    }, { x   } 2 2. {x  . 4/ > y:=. 1 1  }, { x   } 4 4. 8/ > y:= 3*sin(x)+sin(3*x); y := 3 sin( x )  sin( 3 x ). (1/3*(cos(x)^4)+(sin(x))^4); 1 y := cos( x ) 4  sin( x ) 4 3. > y1:=diff(y,x); y1 := 3 cos( x )  3 cos( 3 x ). > y1:=diff(y,x); 4 y1 :=  cos( x ) 3 sin( x )  4 sin( x ) 3 cos( x ) 3. > y2:=solve(y1=0,{x}); 1 1 3 y2 := { x   }, { x   }, { x   } 2 4 4. > y2:=solve(y1=0,{x}); 1 1 { x  0 }, { x    }, { x   } 2 2. 9/. {x  . 1 1  }, { x   } 6 6. > y:= 3*sin(x)-sin(3*x); y := 3 sin( x )  sin( 3 x ). > y1:=diff(y,x); y1 := 3 cos( x )  3 cos( 3 x ) > y2:=solve(y1=0,{x}); 1 y2 := { x   }, { x  0 }, { x   } 2. 1 Lop12.net.

<span class='text_page_counter'>(2)</span> 10/. > y:= sin(2*x)/(2+cos(2*x)); y :=. 13/. sin( 2 x ) 2  cos( 2 x ). > y:=(sin(x))^2 + 3*cos(2*x); y := sin( x ) 2  3 cos( 2 x ). > y1:=diff(y,x); y1 := 2 sin( x ) cos( x )  6 sin( 2 x ). > y1:=diff(y,x); cos( 2 x ) 2 sin( 2 x ) 2 y1 := 2  2  cos( 2 x ) ( 2  cos( 2 x ) ) 2. > y2:=simplify(y1); y2 := 10 sin( x ) cos( x ). > y2:=simplify(y1); 2 cos( 2 x )  1 y2 := 2 4  4 cos( 2 x )  cos( 2 x ) 2. > y3:=solve(y2=0,{x}); 1 y3 := { x   }, { x  0 } 2. > y3:=solve(y2=0,{x}); 1 y3 := { x   } 3. 14/. 11/. y := e x sin( x ). > y1:=diff(y,x); y1 := e x ln( e ) sin( x )  e x cos( x ). > y:= cos(2*x)/(2+sin(2*x)); y :=. > y:=e^(x)*sin(x);. cos( 2 x ) 2  sin( 2 x ). > y2:=simplify(y1); y2 := e x ( ln( e ) sin( x )  cos( x ) ). > y1:=diff(y,x); sin( 2 x ) 2 cos( 2 x ) 2 y1 := 2  2  sin( 2 x ) ( 2  sin( 2 x ) ) 2. > y3:=solve(y2=0,{x}); 1  y3 := { x  arctan   } ln (  e) . > y2:=simplify(y1); 2 sin( 2 x )  1 y2 := 2 5  4 sin( 2 x )  cos( 2 x ) 2. 15/. > y3:=solve(y2=0,{x}); 1 5 y3 := { x    }, { x   } 12 12. > y:=e^(x)*cos(x); y := e x cos( x ). > y1:=diff(y,x); y1 := e x ln( e ) cos( x )  e x sin( x ) > y2:=simplify(y1); y2 := e x ( ln( e ) cos( x )  sin( x ) ). > y:=sqrt(. 12/ cos(1*x))+sqrt(sin(x)); y := cos( x )  sin( x ). > y3:=solve(y2=0,{x}); y3 := { x  arctan ( ln( e ) ) }. > y1:=diff(y,x);. 1 cos( x ) 1 sin( x ) 2 y1 :=   2 cos( x ) sin( x ). 16/. > y:=e^(x)*(x-1); y := e x ( x  1 ). > y1:=diff(y,x); y1 := e x ln( e ) ( x  1 )  e x. > y2:=simplify(y1); ( 3/2 ) ( 3/2 ) 1 sin( x )  cos( x ) y2 := 2 cos( x ) sin( x ). > y2:=simplify(y1); y2 := e x ln( e ) x  e x ln( e )  e x. > y3:=solve(y2=0,{x}); 1 3 y3 := { x   }, { x    } 4 4. > y3:=solve(y2=0,{x}); ln( e )  1 y3 := { x  } ln( e ). 2 Lop12.net.

<span class='text_page_counter'>(3)</span> 17/. > y:=(x)*e^(1-x); y := x e. (1  x ). > y1:=diff(y,x); y1 := 2 x e x  x 2 e x ln( e ). > y2:=simplify(y1); (1  x ) (1  x ) y2 := e xe ln( e ). > y2:=solve(y1=0,{x});. > y3:=solve(y2=0,{x}); 1 y3 := { x  } ln( e ). y2 := { x  0 }, { x  2. (x  1). > y1:=diff(y,x); (x  1) (x  1) y1 := 2 x e  x2 e ln( e ). > y1:=diff(y,x); (x  1) (x  1) y1 := e xe ln( e ). > y2:=solve(y1=0,{x}); y2 := { x  0 }, { x  2. > y2:=simplify(y1); (x  1) (x  1) y2 := e xe ln( e ). y := x e. > y1:=diff(y,x); y1 := e. > y:=(x)^2*e^(x^2); y := x 2 e. 2 (x ). > y1:=diff(y,x); y1 := 2 x e. 2 (x ).  2 x3 e. > y2:=simplify(y1); y2 := 2 x e. 2 (x ).  2 x3 e. 2 (x ). 2 (x ). y1 := 2 x e. ln( e ). > y1:=diff(y,x); 1 1 1 y1 :=  2 4x 2. 1 } ln( e ). 2 (x  1). 1 4x. > y2:=solve(y1=0,{x}); y2 := { x  0 } 25 / > y:=(sqrt(4+x^2))-(sqrt(x^2)); y := 4  x 2  x 2 > y1:=diff(y,x);. 2x e. x). ln( e ). (x  1). 3. ( 1 xe 2. 24/ > y:=(sqrt(4+x))+(sqrt(4-x)); y := 4  x  4  x. 2. > y1:=diff(y,x);. 2 (x  1). . ln( e ). > y:=(x)^2*e^(x^2-1); y := x 2 e. ( x). ( x). > y2:=solve(y1=0,{x}); 1 y2 := { x  4 } ln( e ) 2. > y3:=solve(y2=0,{x}); 1 y3 := { x  0 }, { x   }, { x  ln( e ) 20/. 1 } ln( e ). 23/ > y:=(x)*e^(sqrt(x));. > y3:=solve(y2=0,{x}); 1 y3 := { x   } ln( e ) 19/. 1 } ln( e ). 22/ > y:=(x)^2*e^(x-1); (x  1) y := x 2 e. > y:=(x)*e^(x-1); y := x e. 1 } ln( e ). 21/ > y:=(x)^2*e^(x); y := x 2 e x. > y1:=diff(y,x); (1  x ) (1  x ) y1 := e xe ln( e ). 18/. 1 }, { x  ln( e ). y2 := { x  0 }, { x  . y1 :=. ln( e ). > y2:=solve(y1=0,{x});. x. 4x. 2. . > y2:=solve(y1 =0,{x}); x=0. 3 Lop12.net. x x2.

<span class='text_page_counter'>(4)</span> 26/ > y:=(sqrt(4-x^2))+(sqrt(x^2)); y := 4  x 2  x 2. > y2:=solve(y1 =0,{x}); y2 := { x  2 }. > y1:=diff(y,x);. 31/ > y:= 3*sqrt(9-x)+6*sqrt(x+6); y := 3 9  x  6 x  6. y1 := . x 4  x2. x. . x2. > y1:=diff(y,x); 3 1 y1 :=   2 9x. > y2:=solve(y1 =0,{x}); y2 := { x  2 }, { x   2 }. > y2:=solve(y1 =0,{x}); y2 := { x  6 }. x=0 > restart: 27/ > y:= x+(sqrt(2-x^2)); y := x  2  x 2 > y1:=diff(y,x); y1 := 1 . 32/ > y:= 3*sqrt(12-x)+6*sqrt(x+8); y := 3 12  x  6 x  8 > y1:=diff(y,x); 3 1 y1 :=   2 12  x. x 2  x2. 33/ > y:= x*ln(x); y := x ln( x ). > y:= 3*x-5*(sqrt(4+x^2));. > y1:=diff(y,x); y1 := ln( x )  1. y := 3 x  5 4  x 2. > y1:=diff(y,x); y1 := 3 . > y2:=solve(y1 =0,{x}); ( -1 ) y2 := { x  e }. 5x 4  x2. 34/ > y:= ln(x)/x;. > y2:=solve(y1 =0,{x}); 3 y2 := { x  } 2. y :=. > y1:=diff(y,x);. 29/> y:= 6*x-8*(sqrt(4*x-x^2)); y := 6 x  8 4 x  x 2 > y1:=diff(y,x); y1 := 6 . y1 :=. 4 x  x2. 1 ln( x )  x2 x2. 35/ > y:= ln(x)/x^2; ln( x ) y := x2 > y1:=diff(y,x);. > y:= 3*sqrt(4-x)+6*sqrt(x+6);. y1 :=. y := 3 4  x  6 x  6. > y1:=diff(y,x); 3 1 y1 :=   2 4x. ln( x ) x. > y2:=solve(y1 =0,{x}); y2 := { x  e }. 4 (4  2 x). > y2:=solve(y1 =0,{x}); 4 y2 := { x  } 5 30/. 3 x8. > y2:=solve(y1 =0,{x}); y2 := { x  8 }. > y2:=solve(y1 =0,{x}); y2 := { x  1 } 28/. 3 x6. 1 2 ln( x )  x3 x3. > y2:=solve(y1 =0,{x}); ( 1/2 ) y2 := { x  e }. 3 x6. 4 Lop12.net.

<span class='text_page_counter'>(5)</span> 36/ > y:= ln(x)*x^2; y := ln( x ) x 2. 41/ > y:=1*x-1*ln(5*x^2-10*x+10); y := x  ln( 5 x 2  10 x  10 ). > y1:=diff(y,x); y1 := x  2 x ln( x ). > y1:=diff(y,x);. > y2:=solve(y1 =0,{x}); ( -1 /2 ) y2 := { x  e }. y1 := 1 . 10 x  10 5 x  10 x  10 2. 37/ > y:=( x^2)*ln(x^2); y := x 2 ln( x 2 ). > y2:=solve(y1=0,{x}); y2 := { x  2 }, { x  2 }. > y1:=diff(y,x); y1 := 2 x ln( x 2 )  2 x. 42/ > y:=1*x-1*ln(1*x^2-10*x+10); y := x  ln( x 2  10 x  10 ). > y2:=solve(y1 =0,{x}); y2 := { x   e. ( -1 ). }, { x . e. ( -1 ). > y1:=diff(y,x);. }. y1 := 1 . 38/ > y:=1*x-ln(x^2-1*x+1); y := x  ln( x 2  x  1 ). > y2:=solve(y1=0,{x}); y2 := { x  10 }, { x  2 }. > y1:=diff(y,x);. 43/ > y:=1*x*e^(-x^2/2);. 2x1 y1 := 1  2 x x1. y := x e. > y2:=solve(y1=0,{x}); y2 := { x  2 }, { x  1 }. y1 := 1 . y1 := e. 2 ( 10 x  5 ) 5 x 2  5 x  10.  x2 e. 2 (  1/2 x ). ln( e ). 1 } ln( e ). 44/ > y:=tan(x)+cot(x); y := tan ( x )  cot ( x ). > y1:=diff(y,x); y1 := tan ( x ) 2  cot ( x ) 2. 40/ > y:=1*x+2*ln(5*x^2+5*x+10); y := x  2 ln( 5 x 2  5 x  10 ). y1 := 1 . 2 (  1/2 x ). > y2:=solve(y1=0,{x}); 1 y2 := { x  }, { x   ln( e ). > y2:=solve(y1=0,{x}); y2 := { x  0 }, { x  3 }. > y1:=diff(y,x);. 2 (  1/2 x ). > y1:=diff(y,x);. 39/ > y:=1*x-2*ln(5*x^2+5*x+10); y := x  2 ln( 5 x 2  5 x  10 ). > y1:=diff(y,x);. 2 x  10 x  10 x  10 2. > y2:=solve(y1=0,{x}); 1 1 y2 := { x   }, { x    } 4 4. 2 ( 10 x  5 ) 5 x 2  5 x  10. > y2:=solve(y1=0,{x}); y2 := { x  -1 }, { x  -4 }. 5 Lop12.net.

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