If y= log ( (1-x2/1+x2)) then (dy/dx) is equal to

Question

If y= log ( (1-x2/1+x2)) then (dy/dx) is equal to  (A)  (-4x/1-x4) (B)  (4x3/1-x4) (C)  (1/4-x4) (D)  (-4x3/1-x4)

Tardigrade Admin · Accepted Answer

Given, y = log ((1-x2/1+x2)) = log (1-x2)- log (1+x2) ∴ (d y/d x) =(1/1-x2)(-2 x)-(1/1+x2)(2 x) =-2 x[(1/1-x2)+(1/1+x2)] =-2 x[(1+x2+1-x2/1-x4)]=(-4 x/1-x4)

Q. If $y = lo g (\frac{1 - x ^{2}}{1 + x ^{2}})$ then $\frac{d y}{d x}$ is equal to _______

$\frac{- 4 x}{1 - x ^{4}}$

$\frac{4 x ^{3}}{1 - x ^{4}}$

$\frac{1}{4 - x ^{4}}$

$\frac{- 4 x ^{3}}{1 - x ^{4}}$

Q. If y=log(1+x21−x2​) then dxdy​ is equal to _______

1−x4−4x​

1−x44x3​

4−x41​

1−x4−4x3​

Given, y=log(1+x21−x2​) =log(1−x2)−log(1+x2) ∴dxdy​=1−x21​(−2x)−1+x21​(2x) =−2x[1−x21​+1+x21​] =−2x[1−x41+x2+1−x2​]=1−x4−4x​

Q. If $y = lo g (\frac{1 - x ^{2}}{1 + x ^{2}})$ then $\frac{d y}{d x}$ is equal to _______

$\frac{- 4 x}{1 - x ^{4}}$

$\frac{4 x ^{3}}{1 - x ^{4}}$

$\frac{1}{4 - x ^{4}}$

$\frac{- 4 x ^{3}}{1 - x ^{4}}$