Aug 8, 2018 · #"differentiate using the "color(blue)"chain rule"# #"given "y=f(g(x))" then"# #dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

Dec 10, 2015 · This follows from the definition of natural logarythm and its inverse. ln e^x = x d/dx ln e^x = d/dx x 1/e^x d/dx e^x = 1 d/dx e^x = e^x

# d/dx e^x=e^x# This special exponential function with Euler's number, #e#, is the only function that remains unchanged when differentiated. Method 2 - Power Series. We can use the power series: # e^x = 1 +x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... # Then we can differentiate term by term using the power rule:

Aug 6, 2015 · y^' = (4e^(2x))/(e^(2x) + 1)^2 You can differentiate this function by using the quotient rule and the chain rule.

Mar 10, 2018 · #"differentiate "e^(-x)" using the "color(blue)"chain rule"# #"Given "y=f(g(x))" then"# #dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

Jun 29, 2012 · Homework Statement Proof d/dx e^x=e^x, use e=limit (1+1/h)^h h->infinity Show how that implies d/dx e^x=e^x t Homework Equations The Attempt at a Solution Ive tried using chain rule--wasnt accepted Also, I did e=(1+h)^[1/h]- …

Apr 4, 2018 · Additionly, the number 2.718281 ..., which we call Euler's number) denoted by e is extremely important in mathematics, and is in fact an irrational number (like pi and sqrt(2), And so: d/dx e^x=e^x This special exponential function with Euler's number, e, is the only function that remains unchanged when differentiated.

Dec 9, 2008 · Yup I think I know f(g(x)) but are u using this to explain d/dx e^-x = e^-x and before that, is d/dx e^-x = e^-x correct? e -x is just a very simple example of f[g(x)] . To be asking this question you must have done some calculus lessons before.

Aug 5, 2014 · This is a type of problem involving the product rule. The product rule states: d/dx[f(x) * g(x)] = f'(x)g(x) + f(x)g'(x) So, we will let f(x) = e^x, and g(x) = cos x. We know that the derivative of e^x is simply e^x, and that the derivative of cos x is equal to -sin x. (if these identities look unfamiliar to you, I may recommend viewing videos from …

Sep 16, 2017 · dy/dx = x^(e^(x^2)-1).e^(x^2)(2x^2lnx+1) y = x^(e^(x^2)) ln y = e^(x^2)lnx Apply implicit diferentiation 1/ydy/dx = d/dx(e^(x^2)lnx) = e^(x^2).d/dx lnx + d/dx e^(x^2 ...