Question

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

Answer 1

`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)`