Find the derivative using first principles? : #y=e^(2x)#

1 Answer
Jan 5, 2018

# dy/dx = 2e^(2x) #

Explanation:

Using the limit definition of the derivative:

# y=f(x) => dy/dx = lim_(h rarr 0) (f(x+h)-f(x))/h #

So if #y=e^(2x)#; then:

# dy/dx = lim_(h rarr 0) ( e^(2(x+h)) - e^(2(x)))/h #
# \ \ \ \ \ \ = lim_(h rarr 0) ( e^(2x+2h) - e^(2x))/h #
# \ \ \ \ \ \ = lim_(h rarr 0) ( e^(2x)e^(2h) - e^(2x))/h #
# \ \ \ \ \ \ = lim_(h rarr 0) ( e^(2x)(e^(2h) - 1))/h #
# \ \ \ \ \ \ = e^(2x) \ lim_(h rarr 0) ( e^(2h) - 1)/h #
# \ \ \ \ \ \ = e^(2x) \ lim_(h rarr 0) ( 2(e^(2h) - 1))/(2h) #
# \ \ \ \ \ \ = 2e^(2x) \ lim_(h rarr 0) ( e^(2h) - 1)/(2h) #

Then if we perform, a substitution, #alpha=2h# then clearly:

# h rarr 0 => alpha rarr 0 #

So we can write the derivative as:

# dy/dx = 2e^(2x) \ lim_(alpha rarr 0) ( e^(alpha) - 1)/(alpha) #

Now # lim_(alpha rarr 0) ( e^alpha - 1 ) / alpha = 1# is a standard calculus limit whose limit is unity, and so we find that:

# dy/dx = 2e^(2x) #