If y= sin -1(3x-4x3)+ cos -1(4x3-3x) + tan -1(e), then (dy/dx) is equal to

Question

If y= sin -1(3x-4x3)+ cos -1(4x3-3x) + tan -1(e), then (dy/dx) is equal to (A)  5  (B)  0  (C)  (2/√1+x2)  (D)  (2/√1-x2)

Tardigrade Admin · Accepted Answer

Given, y= sin -1(3x-4x)2+ cos -1(4x3-3x) + tan -1(E) ⇒ y=3 sin -1x+3 cos -1x+ tan -1(E) ⇒ y=(3π /2)+ tan -1(E) On differentiating w.r.t. x, we get (dy/dx)=0

Q. If $y = sin^{- 1} (3 x - 4 x^{3}) + cos^{- 1} (4 x^{3} - 3 x)$ $+ tan^{- 1} (e),$ then $\frac{d y}{d x}$ is equal to

$5$

$0$

$\frac{2}{1 + x ^{2}}$

$\frac{2}{1 - x ^{2}}$

$\frac{5}{1 - x ^{2}}$

Given, $y = sin^{- 1} (3 x - 4 x)^{2} + cos^{- 1} (4 x^{3} - 3 x)$ $+ tan^{- 1} (E)$
$\Rightarrow$ $y = 3 sin^{- 1} x + 3 cos^{- 1} x + tan^{- 1} (E)$
$\Rightarrow$ $y = \frac{3 π}{2} + tan^{- 1} (E)$
On differentiating w.r.t. $x,$ we get $\frac{d y}{d x} = 0$

Q. If y=sin−1(3x−4x3)+cos−1(4x3−3x) +tan−1(e), then dxdy​ is equal to

5

0

1+x2​2​

1−x2​2​

1−x2​5​

Given, y=sin−1(3x−4x)2+cos−1(4x3−3x) +tan−1(E) ⇒ y=3sin−1x+3cos−1x+tan−1(E) ⇒ y=23π​+tan−1(E) On differentiating w.r.t. x, we get dxdy​=0

Q. If $y = sin^{- 1} (3 x - 4 x^{3}) + cos^{- 1} (4 x^{3} - 3 x)$ $+ tan^{- 1} (e),$ then $\frac{d y}{d x}$ is equal to

$5$

$0$

$\frac{2}{1 + x ^{2}}$

$\frac{2}{1 - x ^{2}}$

$\frac{5}{1 - x ^{2}}$

Given, $y = sin^{- 1} (3 x - 4 x)^{2} + cos^{- 1} (4 x^{3} - 3 x)$ $+ tan^{- 1} (E)$
$\Rightarrow$ $y = 3 sin^{- 1} x + 3 cos^{- 1} x + tan^{- 1} (E)$
$\Rightarrow$ $y = \frac{3 π}{2} + tan^{- 1} (E)$
On differentiating w.r.t. $x,$ we get $\frac{d y}{d x} = 0$