Quote from: Shantel on January 30, 2014, 09:25:14 AM
Huh 
Calculus.
The starting function is y = 5e^(x^2). To see that visually, uhh...
http://www.quickmath.com/webMathematica3/quickmath/equations/solve/advanced.jsp#c=solve_advancedsolveequations&v1=y+%3D+5e%5E(x%5E2)&v2=yIt's that. The ^ symbol means we're setting up an exponent.
After that, we're trying to take derivatives. They're Calculus and all, but derivatives are fancy way of saying "what is the rate of change" or "slope" for the function. Since most functions aren't lines, the "slope" of your graph changes as x changes.
Functions that look like y = x^n have pretty nice derivatives. You multiply the right-hand side by the exponent and then reduce the exponent by one. Algebraically, the derivative is y' = n * x^(n - 1). In this problem, you're first using y = x^2, which will have a derivative of y' = 2x.
y = e^x has a wonderful derivative: y' = e^x. Yes, the derivative is the original function.
Multiplying a function by a constant number (like 5) just means your derivative gets multiplied by the same amount. So y = 3x^5 ends up having a derivative of y' = 3*(5x^4) = 15x^4.
Now things get a little weird, but we're almost ready. The chain rule says that if you have a function of a function, like y = f(g(x)), the derivative is y' = f'(g(x))*g'(x). So you first only take the derivative of the outer part and leave the inner intact, and then you multiply the result by the derivative of the inside.
Here we go. Our function is: y = 5e^(x^2). This is a "function of a function" situation. That x^2 means you have an inner function of y = x^2, so let's call it g(x) = x^2. We can basically change our function to now be y = 5e^g(x). If we use f(x) = 5e^x as our outer function, this leads to f(g(x)) = 5e^g(x) = 5e^(x^2).
We need to figure out f'(g(x)), so we first find f'(x). Well, f(x) = 5e^x, so f'(x) = 5e^x (the 5 just copies down, and e^x just goes to e^x). Awesome. Since f and f' are the same, it means f'(g(x)) is just f(g(x)) all over again, so f'(g(x)) = 5e^(x^2).
Next, we need g'(x). g(x) = x^2, so g'(x) = 2x (multiply by 2, lower the exponent from 2 to 1).
Finally we put it all together. [f(g(x))]' = f'(g(x))*g'(x) = 5e^(x^2) * 2x = 10x^(x^2).
Yay, we did math. \o/
[The second derivative isn't much harder, but we did a good job already with the first one.]
