If x loge( loge x) - x2 + y2 = 4(y 0), then dy/dx at x = e is equal to :

Question

If x loge( loge x) - x2 + y2 = 4(y > 0), then dy/dx at x = e is equal to : (A) (e/√4+e2) (B) ((1 + 2e)/ 2 √4+e2) (C) (( 2e - 1 )/ 2 √4+e2) (D) ((1 + 2e)/ √4+e2)

Tardigrade Admin · Accepted Answer

Differentiating with respect to x, x. (1/ℓ nx) . (1/x) +ℓ n(ℓ nx)-2x+2y . (dy/dx) =0 at x = e we get 1-2e+2y (dy/dx) =0 ⇒ (dy/dx) = (2e-1/2y) ⇒ (dy/dx) = (2e-1/2√4+e2) y=(e) = √4+e2

Q. If $x lo g_{e} (lo g_{e} x) - x^{2} + y^{2} = 4 (y > 0)$ , then $d y / d x$ at $x = e$ is equal to :

$\frac{e}{4 + e ^{2}}$

$\frac{( 1 + 2 e )}{2 4 + e ^{2}}$

$\frac{( 2 e - 1 )}{2 4 + e ^{2}}$

$\frac{( 1 + 2 e )}{4 + e ^{2}}$

Q. If xloge​(loge​x)−x2+y2=4(y>0), then dy/dx at x=e is equal to :

4+e2​e​

24+e2​(1+2e)​

24+e2​(2e−1)​

4+e2​(1+2e)​

Differentiating with respect to x, x.ℓnx1​.x1​+ℓn(ℓnx)−2x+2y.dxdy​=0 at x=e we get 1−2e+2ydxdy​=0⇒dxdy​=2y2e−1​ ⇒dxdy​=24+e2​2e−1​y=(e)=4+e2​

Q. If $x lo g_{e} (lo g_{e} x) - x^{2} + y^{2} = 4 (y > 0)$ , then $d y / d x$ at $x = e$ is equal to :

$\frac{e}{4 + e ^{2}}$

$\frac{( 1 + 2 e )}{2 4 + e ^{2}}$

$\frac{( 2 e - 1 )}{2 4 + e ^{2}}$

$\frac{( 1 + 2 e )}{4 + e ^{2}}$