Question

Find the derivative using first principles? : `e^sinx`

Answer 1

`d/dx( e^sinx) = e^sinx cosx`

Explanation:

The definition of the derivative of `y=f(x)` is

`f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`

So Let `f(x) = e^sinx` then the derivative of `y=f(x)` is given by:

`f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`
`" " = lim_(h rarr 0) ( e^sin(x+h)-e^sinx )/h`
`" " = lim_(h rarr 0) ( e^sinx ( e^(sin(x+h)-sinx) - 1 ) ) / h`
`" "= e^sinx lim_(h rarr 0) ( e^(sin(x+h)-sinx) - 1 ) / h`

`" " = e^sinx lim_(h rarr 0) ( e^(sin(x+h)-sinx) - 1 ) / h * (sin(x+h)-sinx)/(sin(x+h)-sinx)`

`" " = e^sinx lim_(h rarr 0) ( e^(sin(x+h)-sinx) - 1 ) / (sin(x+h)-sinx) * (sin(x+h)-sinx)/h`

`" " = e^sinx * L_1 * L_2`

Where:

`L_1 = lim_(h rarr 0) ( e^(sin(x+h)-sinx) - 1 ) / (sin(x+h)-sinx)`

`L_2 = lim_(h rarr 0) (sin(x+h)-sinx)/h`

Let us examine the first limit, `L_1`.
Let `alpha=(sin(x+h)-sinx)` then `alpha rarr 0` as `h rarr 0` and so

`L_1 = lim_(h rarr 0) ( e^(sin(x+h)-sinx) - 1 ) / (sin(x+h)-sinx)`
`\ \ \ \ = lim_(alpha rarr 0) ( e^alpha - 1 ) / alpha`

Now `lim_(alpha rarr 0) ( e^alpha - 1 ) / alpha = 1` is a standard calculus limit and so

`L_1 = 1`

Next we examine the second limit, `L_2`, We can use the sum and product formula:

`sin A - sin B = 2 cos((A + B)/2) sin ((A - B)/2)`

And we get:

`L_2 = lim_(h rarr 0) (sin(x+h)-sinx)/h`
`\ \ \ \ = lim_(h rarr 0) (2cos((x+h+x)/2) sin ((x+h-x)/2))/h`
`\ \ \ \ = lim_(h rarr 0) (2cos((2x+h)/2) sin (h/2))/h`
`\ \ \ \ = lim_(h rarr 0) (cos(x+h/2) sin (h/2))/(h/2)`
`\ \ \ \ = lim_(h rarr 0) cos(x+h/2) * lim_(h rarr 0) (sin (h/2))/(h/2)`

Let `beta = h/2` then `beta rarr 0` as `h rarr 0`, so:

`L_2 = lim_(h rarr 0) cos(x+h/2) * lim_(beta rarr 0) (sin (beta))/(beta)`

And `lim_(beta rarr 0) (sin (beta))/(beta) =1` is another standard calculus limit, giving us:

`L_2 = lim_(h rarr 0) cos(x+h/2) * 1`
`\ \ \ \ = cos(x)`

Combining our results for `L_1` and `L_2` with our earlier result gives us:

`f'(x)=e^sinx * L_1 * L_2`
`" "=e^sinx * 1 * cosx`
`" "=e^sinx cosx`

Answer 2

This is hardly from first principles but it is interesting.....

Explanation:

Start with the definition `e^z = 1 + z + z^2/(2!) + z^3/(3!)...`

We can see that:

`e^(sin x) = 1 + sin x + 1/(2!) sin^2 x + 1/(3!) sin^3 x ...`

And so:

`d/dx( e^(sin x) )= 0 + cos x + 2/(2!) sin x cos x + 3/(3!) sin^2 x cos x + ...`

`= cos x( 1 + sin x + 1/(2!) sin^2 x + ...)`

`= cos x * e^(sin x)`