Question

If $y=1+t^4$ and $x=3t^3+t$ then what is $\frac{dy}{dx}$

• A. $\frac{4t^3}{9t^2+1}$
• B. $\frac{9t^2+1}{4t^3}$
• C. $4t^3(9t^2+1)$
• D. $9t^2-4t^3$

Answer 1

The correct option is A $\frac{4t^3}{9t^2+1}$

Note that the function is given in parametric form. Here differentiating y with respect to x is not straight forward. For this reason we will use chain rule.
$\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{dy}{dt}/\frac{dx}{dt}$
So we differentiate y and x separately with respect to t and divide to get $\frac{dy}{dx}$ as above.
$y=1+t^4$
i.e.,$\frac{dy}{dt}=4t^3\dots \dots \left(1\right)$
$x=3t^3+t$
$\frac{dx}{dt}=9t^2+1\dots \dots \left(2\right)$
$\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4t^3}{9t^2+1}$ [using (1) and (2)]
Hence option (a)