Using the differentiation rules and the derivative table, find the derivative of the function y=x^2(x^4-1)^3


y' = 2x (x^4 - 1)^2 (7x^4 - 1)

We can use the Product Rule: if y = f(x)\cdot g(x) with f and g differentiable, then the derivative y' at x is given by

y' = f'(x)g(x) + f(x)g'(x)

  • Identify the functions


    f(x) = x^2\qquad\text{ and }\qquad g(x) = (x^4 - 1)^3 


  • Calculate the derivatives of f and g

    - Derivative of f(x) = x^2: f'(x) = 2x

    - Derivative of g(x) = (x^4 - 1)^3:

    To derive g(x), we use the Chain Rule. We observe that g(x)=(v(x))^3 with inner function v(x)=x^4-1, so the outer function is the power g(\cdot)=(\cdot)^3. The derivative of g is then

    g'(x) = 3 \big(v(x)\big)^2 \cdot v'(x) Since v'(x) = 4x^3,


    substituting v and v' in the formula for g' we find

    g'(x) = 3 (x^4 - 1)^2 \cdot 4x^3 = 12x^3 (x^4 - 1)^2


  • Apply the Product Rule

    Substituting the expressions of f, g, f' and g' in the formula for the derivative of the product we find

    y' = 2x\cdot (x^4 - 1)^3 + x^2\cdot 12x^3 (x^4 - 1)^2


  • Simplify

    We can collect the term

    2x (x^4 - 1)^2: y' = 2x (x^4 - 1)^2 \left[(x^4 - 1) + 6x^4\right]


    or the final result is

    y' = 2x (x^4 - 1)^2 (7x^4 - 1)