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

Question

Elijah Abram · Accepted Answer

wrong answer

Christopher Allers · Answer

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

Emma Aldridge · Answer

undefined To find the derivative of $ \sin(\sqrt{x}) $ using first principles, we use the definition of the derivative:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Let $ f(x) = \sin(\sqrt{x}) $. Then,

$$ f'(x) = \lim_{h \to 0} \frac{\sin(\sqrt{x + h}) - \sin(\sqrt{x})}{h} $$

Using the trigonometric identity $ \sin(A) - \sin(B) = 2\cos\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right) $, we rewrite the numerator:

$$ f'(x) = \lim_{h \to 0} \frac{2\cos\left(\frac{\sqrt{x + h} + \sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x + h} - \sqrt{x}}{2}\right)}{h} $$

As $ h $ approaches 0, $ \sqrt{x + h} $ approaches $ \sqrt{x} $, so we can rewrite:

$$ f'(x) = \lim_{h \to 0} \frac{2\cos\left(\frac{\sqrt{x} + \sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x} - \sqrt{x}}{2}\right)}{h} $$

$$ = \lim_{h \to 0} \frac{2\cos(\sqrt{x})\sin(0)}{h} $$

$$ = \lim_{h \to 0} \frac{2\cos(\sqrt{x}) \cdot 0}{h} $$

$$ = \lim_{h \to 0} 0 $$

$$ = 0 $$

So, the derivative of $ \sin(\sqrt{x}) $ using first principles is 0.