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))#