As we know, to find a gradient in a linear function, we have to calculate the difference in the objective function (usually y) by the difference in the choice variable (commonly x), you may know this as rise over run or something similar. However, this is not as simple when the function is non linear. Firstly what you will need to do is differentiate your function with respect to the choice variable. An example of this can be seen in the function y=5x2+3x-2 at the point x=3. If we differentiate this function, we get dy/dx= 10x+3. This is from the rule that when you have a variable to a power n (in this case x), you multiply the variable by n and reduce the power by 1. As the variable in the case 3x is to the power of 1, we simply get 3 as our differential, and we get 0 for the differential of -2. This is because -2 is equivalent to -2x0 as x0 is equivalent to 1. Therefore when you multiply -2 by n (0), you get 0. To find the gradient at this point, you simply have to plug the x value given in the question (3) into your differential and you will get the value 33, your gradient at this point.