Question

Find the derivative using first principles? : `sin sqrt(x)`

Answer 1

wrong answer

Answer 2

wrong answer
`d/dx sin sqrt(x) = (cos sqrtx)/(2sqrt(x))`

Explanation:

By definition of the derivative:

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

So with `f(x) = sin sqrt(x)` we have;

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

Using `sin A - sin B -= 2 sin(1/2(A - B)) cos (1/2(A + B))` we can reqrite this as;

`f'(x)=lim_(h rarr 0) ( 2 sin(1/2(sqrt(x+h) - sqrt(x))) cos (1/2(sqrt(x+h) + sqrt(x))) ) / h`

We now use a little trick as ;

`(sqrt(x+h) - sqrt(x))(sqrt(x+h) + sqrt(x)) = (sqrt(x+h))^2-(sqrt(x))^2`
`" " = x+h-x`
`" " = h`

And so we can replace `h` in the denominator as follows:

`f'(x)=lim_(h rarr 0) ( 2 sin(1/2(sqrt(x+h) - sqrt(x))) cos (1/2(sqrt(x+h) + sqrt(x))) ) / ((sqrt(x+h) - sqrt(x))(sqrt(x+h) + sqrt(x)))`
`\ \ =lim_(h rarr 0) ( sin(1/2(sqrt(x+h) - sqrt(x))) cos (1/2(sqrt(x+h) + sqrt(x))) ) / (1/2(sqrt(x+h) - sqrt(x))(sqrt(x+h) + sqrt(x)))`
`\ \ =lim_(h rarr 0) ( sin(1/2(sqrt(x+h) - sqrt(x))) ) / (1/2(sqrt(x+h) - sqrt(x))) * cos (1/2(sqrt(x+h) + sqrt(x)))/(sqrt(x+h) + sqrt(x))`
`\ \ =lim_(h rarr 0) ( sin(1/2(sqrt(x+h) - sqrt(x))) ) / (1/2(sqrt(x+h) - sqrt(x))) * lim_(h rarr 0)cos (1/2(sqrt(x+h) + sqrt(x)))/(sqrt(x+h) + sqrt(x))`

Let's look at the first limit;

`lim_(h rarr 0) ( sin(1/2(sqrt(x+h) - sqrt(x))) ) / (1/2(sqrt(x+h) - sqrt(x)))`

If we put `theta=1/2(sqrt(x+h) - sqrt(x))` then `theta rarr 0` as `h rarr 0`, and so;

`lim_(h rarr 0) ( sin(1/2(sqrt(x+h) - sqrt(x))) ) / (1/2(sqrt(x+h) - sqrt(x))) = lim_(theta rarr 0) sintheta/theta`

Which is a standard trig calculus limit, and is equal to unity.

And so now we have:

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