# d[Cos-1((4x3/27)-x)]/dx=?a)3/(9-x2)1/2b)Negative of the abovec)1/(3-x2)1/2d)Negative of the above optionPlz explain..

Arun Kumar IIT Delhi
9 years ago
Hello
$\\\frac{d}{dx}\left(\arccos \left(\frac{4x^3}{27}-x\right)\right) \\=\left(-\frac{1}{\sqrt{1-u^2}}\right)\left(\frac{4x^2}{9}-1\right) \\u=\frac{4x^3}{27}-x \\=-\frac{3\left(4x^2-9\right)}{\sqrt{-\left(9-4x^2\right)^2\left(x^2-9\right)}}$
Thanks & Regards
Arun Kumar
Btech IIT delhi
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]