Q:

Derivative of $f(x) = y^4$

$(x^3 \cdot y^4)' = ?$
$a = x^3$ , $b = y^4$
$$(x^3 \cdot y^4)'= a'\cdot b + a\cdot b' = 3x^2 \cdot y^4 + x^3 \cdot (b')$$
Not sure what the derivative of $y^4$ would be, $(4y^3)$? Since the derivative of $(x\cdot y) = 1\cdot y + x \cdot y'$ and the derivative of $y' = y'$ instead of following the power rule.

A:

by the product and chain rule we get $$(x^3y(x)^4)'=3x^2\cdot y(x)^4+x^3\cdot 4y(x)^3\cdot y'(x)$$

