Question Number 105132 by mohammad17 last updated on 26/Jul/20

Answered by Dwaipayan Shikari last updated on 26/Jul/20

1)∫5^((x^2 +6)) (x+3)dx  =(1/2)∫5^u du        {u=x^2 +6x   (du/dx)=2x+6  =(1/2) (5^u /(log5))+C=(1/(2log5))5^(x^2 +6x) +C  2)∫e^x cosxdx=I  =e^x sinx−∫e^x sinx  =e^x sinx+e^x cosx−∫e^x cosx=I  2I=e^x (sinx+cosx)  I=(e^x /2)(sinx+cosx)+C

Answered by Dwaipayan Shikari last updated on 26/Jul/20

b)lim_(x→0) ((tanx)/x)=((sinx)/x).(1/(cosx))=(x/x).1=1    as sinx→x

Answered by Dwaipayan Shikari last updated on 26/Jul/20

xy+y+1=x  (dy/dx)x+y+(dy/dx)=1  (dy/dx)=((1−y)/(1+x))  2)xy=1  logx+logy=0  (1/x)+(1/y) (dy/dx)=0  (dy/dx)=−(y/x)=−(1/x^2 )

Answered by Dwaipayan Shikari last updated on 26/Jul/20

a)lim_(x→0) ((e^x +e^(−x) −2)/(1−cos2x))=lim_(x→0) ((e^x −e^(−x) )/(2sin2x))=((e^(2x) −1)/(2e^x sin2x))                                                                     =(1/(4x)).((e^(2x) −1)/e^x )=(1/2)

Answered by Dwaipayan Shikari last updated on 26/Jul/20

(d/dx)(sin((√(1+cos10x))))=(d/dx)(sin((√2)cos5x)=5(√2)cos((√2)cos5x)(−sin5x)

Answered by mathmax by abdo last updated on 26/Jul/20

1)let f(x) =sin((√(1+cos(10x)))) ⇒f(x) =sin((√2)cos(5x)) ⇒  f^′ (x) =−5(√2)sin(5x)cos((√2)cos(5x))

Answered by mathmax by abdo last updated on 26/Jul/20

2) let g(x) = 3 arcsin((x/3))+(√(9−x^2 )) ⇒  g^′ (x) =3×(1/(3(√(1−(x^2 /9))))) +((−2x)/(2(√(9−x^2 )))) =(3/(√(9−x^2 ))) −(x/(√(9−x^2 ))) =((3−x)/(√(9−x^2 )))

Answered by mathmax by abdo last updated on 26/Jul/20

I = ∫ 5^(x^2 +6x) (x+3)dx  changement x+3 =t give I =∫ 5^((t−3)^2  +6(t−3))  t dt  =∫ t  5^(t^2 −6t +9+6t−18)  dt =∫ t 5^(t^2 −9)  dt  =∫ t e^(ln5(t^2 −9))  dt  =(1/(2ln5)) e^(ln5(t^2 −9))  +C =(1/(2ln5))×5^(t^2 −9)  +c =(1/(2ln5))×5^((x+3)^2 −9)  +C

Answered by 1549442205PVT last updated on 26/Jul/20

3a)Lim_(x→0) ((e^x +e^(−x) −2)/(1−cos(2x)))=Lim_(x→0) ((e^x −e^(−x) )/(2sin(2x)))=  =Lim_(x→0) ((e^x +e^x )/(4cos2x))=(2/4)=(1/2)  b)Lim_(x→0) ((tanx)/x)=Lim_(x→0) ((sinx)/x)×Lim_(x→0) (1/(cosx))=1×1=1

Answered by 1549442205PVT last updated on 26/Jul/20

1a)Set y=sin(√(1+cos(10x)))⇒y^2 =sin^2 (√(1+cos(10x)))  Derivative both two sides of above  equation by x we get  2y.y′=2sin(√(1+cos(10x))).((√(1+cos(10x))))′  =2sin(√(1+cos(10x))).(((1+cos(10x))′)/(2(√(1+cos(10x)))))  =2sin(√(1+cos(10x))).((−10sin(10x))/(2(√(1+cos(10x)))))  =−10sin(√(1+cos(10x))).((sin(10x))/(√(1+cos(10x))))  ⇒(dy/dx)=y′=((−10sin(√(1+cos(10x))).((sin(10x))/(√(1+cos(10x)))))/(2y))  =−((5sin(10x))/(2(√(1+cos(10x)))))=((−5sin(10x))/(2(√(2cos^2 (5x)))))  =((−5sin5xcos5x)/((√(2 )) ∣cos5x∣))= { ((((−5(√2))/2)sin5x if x∈(((−π+4kπ)/(10));((π+4kπ)/(10))))),((((5(√2))/2)sin5x if x∈([((π+4kπ)/(10));((3π+4kπ)/(10))))) :}  1b)Set y=3sin^(−1) ((x/3))+(√(9−x^2 ))   y′=(3/(√(1−(x^2 /9))))×(1/3)+((−x)/(√(9−x^2 )))=((3−x)/(√(9−x^2 )))

Answered by 1549442205PVT last updated on 26/Jul/20

II−2)F=∫e^x cosxdx=∫cos xde^x =e^x cos x+∫e^x sinxdx  =e^x cosx+∫sinxde^x =e^x cosx+e^x sinx−∫e^x cosxdx  F=e^x (cosx+sinx)−F⇒2F=e^x (cosx+sinx)  ⇒F=((e^x (cosx+sinx))/2)+C  II−1)∫5^(x^2 +6x) (x+3)=(1/2)∫5^(x^2 +6x) d(x^2 +6x)  =(5^(x^2 +6x) /(2ln5))+C

Answered by 1549442205PVT last updated on 26/Jul/20

4a)xy+y+1=x⇔y(x+1)+1=x⇒y=((x−1)/(x+1))=1−(2/(x+1))  y′=(2/((x+1)^2 ))  b)xy=1⇒y=(1/x)⇒y′=−(1/x^2 )