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

2 Answers
Feb 9, 2017

wrong answer

Feb 9, 2017

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))