y = 3t^5

x =5t^3.

Here both x and y are expressed in terms of t. This is called the parametric equation of the curve .

Therefore dy/dx = (dy/dt)* dt/dx. . Or

dy/dx = (dy/dt)/(dx/dt)

y = 3t^5. So dy/dt = (3t^5)' = 3*5*t^(5-1) = 15t^4.

x =5t^3 . So dx/dt = (5t^3)' = 5*3t^(3-1) = 15t^2.

Therefore dy/dx = (dy/dt)/(dx/dt) = 15t^4/(15t^2) = t^2.

Therefore dy/dx = t^2 .

The answer posted above in incomplete to the extent that the value of dy/dx is given in terms of t rather than x.

We can complete the answer by convering the value of dy/dt in terms of x.

It is given:

x = 5t^3

Therefor:

t = (x/5)^(1/3)

Substituting this value of t in the value of dy/dx we get:

dy/dx = t^2 = [(x/5)^(1/3)]^2

= (x/5)^(2/3)

We notice that x and y are expressions that contain the variable t.

We'll differentiate x and y with respect to t.

dy/dx = (dy/dt)/(dx/dt)

x=5t^3 and y=3t^5

dy/dt = (d/dt)(3t^5)

dy/dt = 15t^4 (1)

(dx/dt) = (d/dt)(5t^3)

(dx/dt) = 15t^2 (2)

Now, we'll insert (1) and (2) in the ratio:

dy/dx = 15t^4/15t^2

We'll simplify and we'll get:

**dy/dx = t^2**