How do you find the derivative of $y = {cos}^{3} w + cos (w^{3})$ $y= cos^3 w + cos(w^3)$ ?

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Answer 1 · 2017-03-05T01:17:36

$\frac{d y}{d x} = - 3 sin w {cos}^{2} w - 3 w^{2} sin (w^{3})$ $dy/dx = -3sinwcos^2w - 3w^2sin(w^3)$

$\frac{d}{d x} [a + b + c] = \frac{d}{d x} [a] + \frac{d}{d x} [b] + \frac{d}{d x} [c]$ $d/dx[a+b+c] = d/dx[a] + d/dx[b] + d/dx[c]$

so

$\frac{d y}{d x} = \frac{d}{d x} [{cos}^{3} w] + \frac{d}{d x} [cos (w^{3})]$ $dy/dx = d/dx[cos^3w] + d/dx[cos(w^3)]$

Now we've split it up, we can tackle each term separately.

The product rule states that

$\frac{d}{d x} a b = b \frac{d}{d x} [a] + a \frac{d}{d x} [b]$ $d/dx ab = bd/dx[a] + ad/dx[b]$

so

$\frac{d}{d x} {cos}^{3} w = cos w \frac{d}{d x} [{cos}^{2} w] + {cos}^{2} w \frac{d}{d x} cos w$ $d/dxcos^3w = coswd/dx[cos^2w] + cos^2wd/dxcosw$

$\frac{d}{d x} {cos}^{2} w = cos w \frac{d}{d x} cos x + cos w \frac{d}{d x} cos w$ $d/dxcos^2w = coswd/dxcosx + coswd/dxcosw$

$= - 2 sin w cos w$ $= -2sinwcosw$

therefore,

$\frac{d}{d x} {cos}^{3} w = cos w \cdot - 2 sin w cos w + {cos}^{2} w \cdot - sin w$ $d/dxcos^3w = cosw*-2sinwcosw + cos^2w*-sinw$

$= - 3 sin w {cos}^{2} w$ $= -3sinwcos^2w$

Now we can begin to look at the second term, for which we need the chain rule:

$d/dx f(g(x)) = g'(x)f'(g(x))$

so

$d/dx cos(w^3) = 3w^2 * -sin(w^3) = -3w^2sin(w^3)$

Now we can put the whole thing back together,

$dy/dx = -3sinwcos^2w-3w^2sin(w^3)$

How do you find the derivative of y= cos^3 w + cos(w^3)y=cos3w+cos(w3)y= cos^3 w + cos(w^3)?