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

Question

Christopher Agresta · Accepted Answer

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

#d/dx[a+b+c] = d/dx[a] + d/dx[b] + d/dx[c]#

so

#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

#d/dx ab = bd/dx[a] + ad/dx[b]#

so

#d/dxcos^3w = coswd/dx[cos^2w] + cos^2wd/dxcosw#

#d/dxcos^2w = coswd/dxcosx + coswd/dxcosw#

#= -2sinwcosw#

therefore,

#d/dxcos^3w = cosw*-2sinwcosw + cos^2w*-sinw#

#= -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)#

Christopher Allers · Answer

undefined To find the derivative of $y = \cos^3(w) + \cos(w^3)$, you would use the chain rule and the derivative of cosine function. The derivative would be:

$$y' = -3\cos^2(w) \sin(w) - \sin(w^3) \cdot 3w^2 \sin(w^3 - 3\cos(w^3) \cdot w^2\sin(w^3 - w^2)$$

Alternatively, you can express the derivative as:

$$y' = -3\cos^2(w) \sin(w) - 3w^2\sin(w^3) \sin(w^3) - 3\cos(w^3)w^2\sin(w^3 - w^2)$$