y = f(x) = 5(x^2-3x +4)^5

Question

Find the derivative of the following function:

y = f(x) = 5(x^2-3x +4)^5. Hint use the Chain Rule.

Answer 1

y = f(x) = 5(x^2-3x+4)^5

Apply derivative with respect of x .

= d/dx 5(x^2-3x+4)^5

Apply chain rule and d/dx(x^n) = nx^n-1

= 5*5(x^2-3x+4)^5-1*d/dx(x^2-3x+4)

= 25(x^2-3x+4)^4*d/dx(x^2)-d/dx(3x)+d/dx(4)

= 25(x^2-3x+4)(2x-3)