How do you use the product Rule to find the derivative of #(2x)(-sinx) + (cosx)(2) - 2(cosx)#?

Question

Elijah Abram · Accepted Answer

First of all, notice how this cancels to give:

#(2x)(-sinx)#

Much easier now. Using the product rule, you have:

#f(x) = g(x)h(x)#

#(df)/(dx) = g(x)(dh)/(dx) + h(x)(dg)/(dx)#

So, you get:

#= (2x)*d/(dx)[-sinx] + (-sinx)*d/(dx)[2x]#

#= (2x)*(-cosx) + (-sinx)*2#

#= color(blue)(-2xcosx - 2sinx)#

Emily Ammerman · Answer

undefined To find the derivative using the product rule, apply the formula:

$$ \frac{d}{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x) $$

Given $ f(x) = 2x $ and $ g(x) = -\sin(x) $, the derivatives are $ f'(x) = 2 $ and $ g'(x) = -\cos(x) $.

$$ \frac{d}{dx} [2x(-\sin(x))] = (2)(-\sin(x)) + (2x)(-\cos(x)) $$

Given $ f(x) = \cos(x) $ and $ g(x) = 2 $, the derivatives are $ f'(x) = -\sin(x) $ and $ g'(x) = 0 $.

$$ \frac{d}{dx} [\cos(x)(2)] = (-\sin(x))(2) + (\cos(x))(0) $$

Finally, given $ f(x) = -2 $ and $ g(x) = \cos(x) $, the derivatives are $ f'(x) = 0 $ and $ g'(x) = -\sin(x) $.

$$ \frac{d}{dx} [-2\cos(x)] = (0)(\cos(x)) + (-2)(-\sin(x)) $$

Summing up these results, we get the derivative:

$$ \frac{d}{dx} [2x(-\sin(x)) + \cos(x)(2) - 2\cos(x)] = -2\sin(x) + 2x(-\cos(x)) - 2\sin(x) $$