Derivatives of f(x) = u 

Considering f(x) = (2  5x)^{3}, find f'(x) Let’s rewrite the function as follows: y = (5x + 2)^{3} Let's consider the function u = 5x + 2 u' = 5 We know that the derivative of u^{n}, f'(u) = nu^{n1}du/dx. Therefore: u^{n} = (5x + 2)^{3} (u^{n})' = nu^{n1}u' = 3(5x + 2)^{2}(5) = 15(5x + 2)^{2} Considering f(x) = (7  2x^{3})^{4}, find f'(x) Let’s rewrite the function as follows: y = (2x^{3} + 7)^{4} Let's consider the function u as follows u = 2x^{3} + 7 u' = 6x^{2} Let's consider the function u as follows f = u^{4} We know that the derivative of u^{n}, f'(u) = nu^{n1}du/dx. Therefore: u^{n} = (2x^{3} + 7)^{4} (u^{n})' = nu^{n1}u' = 4(2x^{3} + 7)^{5}(6x^{2}) = 24x^{2}(7  2x^{3})^{5}
Let's consider u = 4x  7 u' = 4 Remember the formula to find the derivative of the inverse of a function:
Let's consider u = x  1 u' = 1 And let's consider v = x + 1 v' = 1 Remember the formula to find the derivative of the division of two functions:
Let’s rewrite the function as follows: y = (2x + 1)^{3} Let's consider the function u = 2x + 1 u' = 2 We know that the derivative of u^{n}, f'(u) = nu^{n1}du/dx. Therefore: u^{n} = (2x + 1)^{3} (u^{n})' = nu^{n1}u' = 3(2x + 1)^{4}(2) = 6(2x + 1)^{4}



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