Using the chain rule, the derivative is given by:
[tex]\frac{df}{dt} = 4t^3 + 6t^5[/tex]
Suppose we have a function f(x,y), with both x and y functions of a variable t, that is:
x = x(t)
y = y(t)
Hence, the derivative of f as a function of t is given by:
[tex]\frac{df}{dt} = \frac{df}{dx}\frac{dx}{dt} + \frac{df}{dy}\frac{dy}{dt}[/tex]
In this problem, we have that:
[tex]f(x,y) = x^2 + y^2[/tex]
[tex]x(t) = t^2[/tex]
[tex]y(t) = t^3[/tex]
Hence:
[tex]\frac{df}{dx} = 2x = 2t^2[/tex]
[tex]\frac{dx}{dt} = 2t[/tex]
[tex]\frac{df}{dy} = 2y = 2t^3[/tex]
[tex]\frac{dy}{dt} = 3t^2[/tex]
Hence:
[tex]\frac{df}{dt} = \frac{df}{dx}\frac{dx}{dt} + \frac{df}{dy}\frac{dy}{dt}[/tex]
[tex]\frac{df}{dt} = 2t^2(2t) + 2t^3(3t^2)[/tex]
[tex]\frac{df}{dt} = 4t^3 + 6t^5[/tex]
More can be learned about the chain rule at https://brainly.com/question/10309252
