Differentiate y = 5x^3 + 7x + 3 with respect to x

When differentiating with respect to x, we multiply the coefficient of the 'x' by the power of the 'x', and subtract one from the power of 'x'. For example:y = 5x^5dy/dx = (55)x^(5-1) = 25x^4If the power of 'x' is then 0, this will mean that we remove the 'x' from the differentiated equation. This is because any number to the power of 0 (e.g. 5^0) will always equal one. Therefore the coefficient of the 'x' will be just one its own. For example:y = 3x = 3x^1dy/dx = 3x^(1-1) = 3x^0 = 3Finally, differentiating a number that is not a coefficient to the variable that we are differentiating with respect to will result in the number being removed from the differentiated result. For example:y = 9x + 3dy/dx = 9 + 0 = 9The 3 has therefore been removed as it was not a coefficient of any 'x' variable.With this knowledge, we can now answer the initial question.{ANSWER: 5x^3 + 7x +3 = (53)x^(3-1) + 7x(1-1) + 0 = 15x^2 + 7x^0 = 15x^2 + 7}