I am utterly amazed how that equates. This computation seems to have generated a Live on its own. Thank you for clarification, sir. It’s greatly appreciated!
@DarklightALBANIA it's the same chain rule, just different notation. In this example, using your notation we have, f(x) = e^x and g(x) = (2x). So f(g(x)) = e^(2x). And the derivative is just what you said, f '(g(x))*g '(x) = e^(2x)*2
You can just put the co-efficient behind the derivative it doesnt make a difference and then take the derivative and multiply by the co-efficient. You can very easily prove it with the product rule: f(x)g(x) = f'(x)g(x) + f(x)g'(x)
I love the way you explain. But it seems that you miss the derivative of natural logarithm and exponential functions part... I couln't find that topics on your Calculus playlist. By the way, great work!
+Serge Espejo e in this case is not a variable, it is a fundamental constant in mathematics. Its approximate value 2.718, but its important property is that the derivative of e^x is e^x. This would not work with any other number.