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Derivative of a product term and a trig term

Let f(x)=x3cos⁡(x)−4tan⁡(x)f(x) = x^3\cos(x) - 4\tan(x). Find f′(x)f'(x).

Hint

Apply the product rule to x3cos⁡(x)x^3\cos(x) (recall ddx[cos⁡x]=−sin⁡x\frac{d}{dx}[\cos x] = -\sin x), and recall that ddx[tan⁡x]=sec⁡2x\frac{d}{dx}[\tan x] = \sec^2 x. Then combine using the constant multiple and difference rules.

Step-by-step solution

First differentiate x3cos⁡(x)x^3\cos(x) using the product rule: ddx[x3cos⁡(x)]=(ddx[x3])cos⁡(x)+x3(ddx[cos⁡(x)])=3x2cos⁡(x)+x3(−sin⁡(x))=3x2cos⁡(x)−x3sin⁡(x)\frac{d}{dx}[x^3\cos(x)] = \left(\frac{d}{dx}[x^3]\right)\cos(x) + x^3\left(\frac{d}{dx}[\cos(x)]\right) = 3x^2\cos(x) + x^3(-\sin(x)) = 3x^2\cos(x) - x^3\sin(x).

Next, differentiate −4tan⁡(x)-4\tan(x) using the constant multiple rule together with ddx[tan⁡x]=sec⁡2x\frac{d}{dx}[\tan x] = \sec^2 x, giving −4sec⁡2(x)-4\sec^2(x). Adding the two results, f′(x)=3x2cos⁡(x)−x3sin⁡(x)−4sec⁡2(x)f'(x) = 3x^2\cos(x) - x^3\sin(x) - 4\sec^2(x).

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