How do you find the derivative of ${sin}^{2} x + {cos}^{2} x$ $sin^2x+cos^2x$ ?

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Answer 1 · 2015-12-19T18:52:19

$\frac{d}{d x} ({sin}^{2} x + {cos}^{2} x) = \frac{d}{d x} (1) = 0$ $d/(dx)(sin^2 x + cos^2 x) = d/(dx)(1) = 0$

The easiest way is to use the Pythagorean identity:

${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x + cos^2x = 1$

Then the derivative of a constant ( $1$ $1$ ) is $0$ $0$ .

Alternatively you can use a combination of the power rule, chain rule and derivatives of $sin x$ $sin x$ and $cos x$ $cos x$ as follows:

Assume we know:

$\frac{d}{d x} sin x = cos x$ $d/(dx) sin x = cos x$

$\frac{d}{d x} cos x = - sin x$ $d/(dx) cos x = -sin x$

Power rule:

$\frac{d}{d x} x^{n} = n x^{n - 1}$ $d/(dx) x^n = n x^(n-1)$

Chain rule:

$d/(dx) u(v(x)) = u'(v(x)) * v'(x)$

Then:

$d/(dx)(sin^2 x + cos^2 x) = 2 sin x cos x - 2 cos x sin x = 0$

How do you find the derivative of sin^2x+cos^2xsin2x+cos2xsin^2x+cos^2x?