Question

If $t(1+x^2)=x$ and $x^2+t^2=y$ then $\frac{dy}{dx}$ at $x=2$ is

• A. $\frac{88}{125}$
• B. $\frac{488}{125}$
• C. $1$
• D. $\frac{244}{125}$

Answer 1

The correct option is B $\frac{488}{125}$

Given:
$t(1+x^2)=x$ and $x^2+t^2=y$
So, $x^2+\frac{x^2}{(1+x^2{)}^2}=y$
Differentiating w.r.t $x$, we get

$2x+\frac{2x(1+x^2{)}^2-4x^3(1+x^2)}{(1+x^2{)}^4}=\frac{dy}{dx}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=2}=4-\frac{12}{125}=\frac{488}{125}$