Question

Find \(\frac{dy}{dx}\), if x=9t^4 and y=t.

(a) \(\frac{1}{36t^3}\)

(b) \(\frac{1}{36t^2}\)

(c) \(\frac{-1}{36t^3}\)

(d) \(\frac{1}{32t^3}\)

I had been asked this question during an online interview.

My query is from Derivatives of Functions in Parametric Forms in division Continuity and Differentiability of Mathematics – Class 12

Answer 1

The correct answer is (a) \(\frac{1}{36t^3}\)

The best I can explain: Given that, x=9t^4 and y=t

\(\frac{dx}{dt}\)=36t^3

\(\frac{dy}{dt}\)=1

∴\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{1}{36t^3}\)