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d[Cos -1 ((4x 3 /27)-x)]/dx=? a)3/(9-x 2 ) 1/2 b)Negative of the above c)1/(3-x 2 ) 1/2 d)Negative of the above option Plz explain..

d[Cos-1((4x3/27)-x)]/dx=?


a)3/(9-x2)1/2


b)Negative of the above


c)1/(3-x2)1/2


d)Negative of the above option


 


Plz explain..

Grade:12

3 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
9 years ago
Hello
Thanks & Regards
Arun Kumar
Btech IIT delhi
Askiitians Faculty
Ruthwik
13 Points
4 years ago
4x^3-27x/27=4(x/3)^3-x____(1)
Let x/3=cos theta
Theta = cos^-1x/3
Multiply and divide (1) with 3
4(x/3)^3-3x/3
4cos^3theta-3costheta=cos3theta
Substitute the value of cos3theta in the given question 
I.e, cos^-1(4×^3-27×/27)
We get cos^-1(cos3theta)
I.e, 3theta=y
Dy/dx= d/dx(3theta) 
3d/dxtheta
Theta=cos^-1 ×/3
3(-1/(1-×^2/9)^1/2
3{-1/(9-×^2/9)^1/2}
 
 
 
Sandeep
103 Points
3 years ago
By substitution method
we can convert (4x3/27-x) of type (4cos3y-3cosy) which is equal to cos3y
so put x/3=cosy ----1
Then it would be
let z= cos-1[4x3/27-x]
from eq--1
z=cos-1[4cos3y-cosy]
z=cos-1[cos3y]
z=3y
as x/3=cosy
y=cos-1(x/3)
z=3cos-1(x/3)
Now
dz/dx =3*dcos-1(x/3)/dx
dcos-1x/dx= -1/(1-x2)½.    ----formula
let t=x/3
By chain rule
dz/dx=3[dcos-1t/dt*dt/dx]
dz/dx= 3[-1/(1+(x/3)1/2 *d(x/3)/dx]
dz/dx= 3* [-3/(9+x2)*1/3]
dz/dx= -3/(9+x2)              ----(answer)
 
 

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