Question

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

• A. $\frac{488}{125}$
• B. $\frac{88}{125}$
• C. $\frac{101}{125}$
• D. None of these

Answer 1

Answer: (A)

$\frac{488}{125}$

$x^2+t^2=y\Rightarrow \frac{dy}{dx}=2x+2t.\frac{dt}{dx}$ ...(1)

As $t=\frac{x}{1+x^2}\Rightarrow \frac{dt}{dx}=\frac{1-x^2}{{(1+x^2)}^2}$

Substitute these value of $t$ and $\frac{dt}{dx}$ in (1), we get

$\frac{dy}{dx}=2x+\frac{2x}{1+x^2}.\frac{1-x^2}{{(1+x^2)}^2}$

On ptiing $x=2$ in $\frac{dy}{dx}$, we get

$\frac{dy}{dx}=\frac{488}{125}$