How do you differentiate $y = {sin}^{2} 4 x + \frac{1}{2} cos 8 x$ $y = sin^2 4x + 1/2 cos 8x$ ?

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How do you differentiate $y = {sin}^{2} 4 x + \frac{1}{2} cos 8 x$ $y = sin^2 4x + 1/2 cos 8x$ ?

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Answer 1 · 2016-11-06T06:50:17

$\frac{d y}{d x} = 0$ $dy/dx=0$ .

Explanation:

$y = {sin}^{2} (4 x) + \frac{1}{2} cos (8 x)$ $y=sin^2(4x)+1/2cos(8x)$
multiplying both sides by $2$ $2$ we get $2 y = 2 {sin}^{2} (4 x) + cos (8 x)$ $2y=2sin^2(4x)+cos(8x)$
we know the rule of trigonometric multiple angles: $2 {sin}^{2} x = 1 - cos (2 x)$ $2sin^2x=1-cos(2x)$
so $2 {sin}^{2} (4 x) = 1 - cos (8 x)$ $2sin^2(4x)=1-cos(8x)$
so the equation reduces to $2 y = 1 - cos (8 x) + cos (8 x)$ $2y=1-cos(8x)+cos(8x)$ = $1$ $1$
differentiating both sides w.r.t $x$ $x$ we get $\frac{d y}{d x} = 0$ $dy/dx=0$ .

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How do you differentiate $y = {sin}^{2} 4 x + \frac{1}{2} cos 8 x$ $y = sin^2 4x + 1/2 cos 8x$ ?

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Explanation:

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How do you differentiate $y = {sin}^{2} 4 x + \frac{1}{2} cos 8 x$ $y = sin^2 4x + 1/2 cos 8x$ ?