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How do you find the derivative of 5x? - Answers

Let y = 5x Then think y = 5x^(1) Differentioation at its most simplistic is to move the exponential to be a coefficient. Then taking the original expoential subtract '1' from it , and use the answer as the new expoential . Algebraically y = Ax^(n) dy/dx = Anx^(n-1) So for the above example y = 5x^(1) dy/dx = 5(1)x^(1-1) dy/dx = 5(1)x^(0) Now any value to the power of '0' equa; '1'. Hence dy/dx = 5(1)(1) dy/dx = 5 The answer.



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How do you find the derivative of 5x? - Answers

https://math.answers.com/calculus/How_do_you_find_the_derivative_of_5x

Let y = 5x Then think y = 5x^(1) Differentioation at its most simplistic is to move the exponential to be a coefficient. Then taking the original expoential subtract '1' from it , and use the answer as the new expoential . Algebraically y = Ax^(n) dy/dx = Anx^(n-1) So for the above example y = 5x^(1) dy/dx = 5(1)x^(1-1) dy/dx = 5(1)x^(0) Now any value to the power of '0' equa; '1'. Hence dy/dx = 5(1)(1) dy/dx = 5 The answer.



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https://math.answers.com/calculus/How_do_you_find_the_derivative_of_5x

How do you find the derivative of 5x? - Answers

Let y = 5x Then think y = 5x^(1) Differentioation at its most simplistic is to move the exponential to be a coefficient. Then taking the original expoential subtract '1' from it , and use the answer as the new expoential . Algebraically y = Ax^(n) dy/dx = Anx^(n-1) So for the above example y = 5x^(1) dy/dx = 5(1)x^(1-1) dy/dx = 5(1)x^(0) Now any value to the power of '0' equa; '1'. Hence dy/dx = 5(1)(1) dy/dx = 5 The answer.

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      Let y = 5x Then think y = 5x^(1) Differentioation at its most simplistic is to move the exponential to be a coefficient. Then taking the original expoential subtract '1' from it , and use the answer as the new expoential . Algebraically y = Ax^(n) dy/dx = Anx^(n-1) So for the above example y = 5x^(1) dy/dx = 5(1)x^(1-1) dy/dx = 5(1)x^(0) Now any value to the power of '0' equa; '1'. Hence dy/dx = 5(1)(1) dy/dx = 5 The answer.

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