# quick differentiation question about sin^2(x)

• June 24th 2010, 08:04 AM
Red
Ignorant question here, but I was looking at this example i have, needing to take the derivative:

f'(x) = sin^2(x) + cos(x)
= 2sin(x)*d/dx(sin x) - sin(x)
= 2sin(x)*cos(x) - sin(x)

and so on... I'm wanting to know why the 2sin(x) isn't where the differentiation stops for sin^2(x)? there's the 2sin(x), and then the d/dx(sinx) added in as well. The cos(x) on the right side doesn't have any extra steps for it. Hope this makes sense, and i realize this is a silly question but I just wanted the clarification.
• June 24th 2010, 08:12 AM
Unknown008
You can consider this function as follows:

$f(x) = (sin(x))^2 + cos(x)$

Then,

$f'(x) = 2(sin(x)).cos(x) - sin(x) = 2sin(x)cos(x) - sin(x)$

Just as the equation:

$y = (x^2+2)^2$

becomes:

$y' = 2(x^2 + 2). (2x) = 4x(x^2 + 2)$
• June 24th 2010, 08:15 AM
Also sprach Zarathustra
$f(x)=sin^2(x)$

$f'(x)=2sin(x)\cdot (sinx)'=2sin(x)\cdot cos(x)=sin(2x)$
• June 24th 2010, 10:43 AM
TheCoffeeMachine
Quote:

Originally Posted by Red
I'm wanting to know why the 2sin(x) isn't where the differentiation stops for sin^2(x)?

Because it's not true that [LaTeX ERROR: Convert failed] . In general, [LaTeX ERROR: Convert failed] unless [LaTeX ERROR: Convert failed] The derivative of f at x is defined as the limit [LaTeX ERROR: Convert failed] , but unfortunately, many students (at school, anyway) understand it as [LaTeX ERROR: Convert failed] . The power rule is itself as a result of considering the function [LaTeX ERROR: Convert failed] at [LaTeX ERROR: Convert failed] , and using the definition of the derivative to calculate [LaTeX ERROR: Convert failed] as [LaTeX ERROR: Convert failed] . If I were to ever teach (school) calculus, it's one of those things I would write in giant letters somewhere above the board. (Lipssealed)
• June 24th 2010, 11:27 AM
Also sprach Zarathustra
You can write only the words: "By definition"