Find the derivative using first principles? : sin sqrt(x)
2 Answers
wrong answer
wrong answer
Explanation:
By definition of the derivative:
f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h
So with
f'(x)=lim_(h rarr 0) ( sinsqrt(x+h)-sin(sqrt(x)) ) / h
Using
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 ;
And so we can replace
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
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))