come calcolare la derivata di (1-3x)^4(1+x)

$$y=(1-3x)^4(1+x)$$

 

Let   $$y1=(1-3x)^4$$   and   $$y2=1+x$$

 

$$\frac{dy1}{dx}=4(1-3x)\times(-3)=-12(1-3x)$$

 

$$\frac{dy2}{dx}=1$$

 

From the product rule:  $$\frac{dy}{dx}=y1\times \frac{dy2}{dx}+y2\times \frac{dy1}{dx}$$

 

So:

$$\frac{dy}{dx}=(1-3x)^4\times 1+(1+x)\times (-12(1-3x))$$

 

or

$$\frac{dy}{dx}=(1-3x)^4-12(1-3x)(1+x)$$

 

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