Question

If $x = \sqrt{1-\tan y}$, then $\frac{dy}{dx} =$

• A. $\frac{2x}{x^4+2x^2+2}$
• B. $-\frac{2x}{x^4-2x^2+2}$
• C. $\frac{2x}{x^4-2x^2+2}$
• D. $-\frac{2x}{x^4+2x^2+2}$

Hint:

When asked to find $\frac{dy}{dx}$ from a relation where $x$ is given in terms of $y$, it's often easier to first algebraically solve for $y$ in terms of $x$ and then perform direct differentiation, rather than using implicit differentiation or finding $\frac{dx}{dy}$ first.

Answer 1

The Correct Option is B

We are given the relation $x = \sqrt{1-\tan y}$.
First, we express $y$ as an explicit function of $x$.
Square both sides: $x^2 = 1-\tan y$.
Rearrange to solve for $\tan y$: $\tan y = 1-x^2$.
Take the inverse tangent of both sides: $y = \arctan(1-x^2)$.
Now, we differentiate $y$ with respect to $x$.
$\frac{dy}{dx} = \frac{d}{dx}\left(\arctan(1-x^2)\right)$.
Using the chain rule and the formula $\frac{d}{du}(\arctan u) = \frac{1}{1+u^2}$:
$\frac{dy}{dx} = \frac{1}{1+(1-x^2)^2} \cdot \frac{d}{dx}(1-x^2)$.
$\frac{dy}{dx} = \frac{1}{1+(1-2x^2+x^4)} \cdot (-2x)$.
Simplify the expression:
$\frac{dy}{dx} = \frac{-2x}{1+1-2x^2+x^4} = \frac{-2x}{x^4-2x^2+2}$.