How do I find the equation of the tangent to y = e^(x^2) at the point x = 4?

This question can be broken up into two main parts, one you're probably familiar with from C1 + C2 and one which is newer. The first part relates to differentiating a compound function (where a function is declared in terms of another function, see C3 chapter on functions if you're still unclear about this term). For this, we must utilise a method of differentiation called the "chain rule". By this method, we can differentiate things given in the form f(g(x)), where here f(x) = e^x and g(x) = x^2. The first step is to differentiate g(x), here this gives us 2x. The second step is to differentiate f(x), giving us e^x (remember e^x is a particularly unique function as its derivative is the function itself). The final step of the chain rule is to put together each separate part into a single function in the form g'(x)f'(g(x)) giving us the answer 2xe^(x^2).

Now we have the differentiated function, the method is the same as finding the tangent of any other curve that you've met in earlier modules. The find the gradient of the tangent at the point x = 4, we must substitute this into our gradient function. This give us f'(4) = 2(4)e^(4^2) = 8e^16. (It is important to remember to always keep this value exact in terms of "e"). As we were not provided with a y-coordinate in the question, we must calculate this by substituting x = 4 into the original equation. This will tell us that y = e^(4^2) = e^16. Now we know both the x and y coordinates, as well as the gradient at the point x = 4, we can deduce the equation of the tangent using the formula (y - y1) = m*(x - x1) [where y1 and x1 are the y and x coordinates found]. Therefore the equation of the tangent will be (y - e^16) = 8e^16(x - 4). Unless asked for in a specific form, your answer can be left exactly as this!