Question

If x = 2t cube and y = 3t square, then value of dy/dx is

Answer 1

have a look at this pic

Answer 2

Answer:

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{6t}{6t^2}=\frac{1}{t}$

Step-by-step explanation:

Given x and y

$x=2t^3, y=3t^2$

$\text{we have to find }\frac{\mathrm{d} y}{\mathrm{d} x}$

$\text{Using }\frac{\mathrm{d} (ax^n)}{\mathrm{d} x}=nax^{n-1}$

$x=2t^3$

$\frac{\mathrm{d} x}{\mathrm{d} t}=2(3t^{3-1})=6t^2$    →  (1)

$y=3t^2$

$\frac{\mathrm{d} y}{\mathrm{d} t}=2(3t^{2-1})=6t$  →   (2)

Divide eq 2 by 1

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{6t}{6t^2}=\frac{1}{t}$