How do you differentiate $f(x)=(x^2+1)(x^2-1)$ using the product rule?

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Answer 1 · 2016-02-27T19:01:35

$d/dx(x^2+1)(x^2-1)=4x^3$

$d/dx(x^2+1)(x^2-1)$

Applying product rule: $(f\cdot g)^'=f^'\cdot g+f\cdot g^'$

$f=x^2+1,g=x^2-1$

$=\frac{d}{dx}(x^2+1)(x^2-1)+\frac{d}{dx}(x^2-1)(x^2+1)$ ...(i)

$\frac{d}{dx}(x^2+1)=2x$

(Applying sum/difference rule : $(f\pm g)^'=f^'\pm g^'$
$=\frac{d}{dx}(x^2)+\frac{d}{dx}(1)$ $=2x+0$ $=2x$ )

Also,
$frac{d}{dx}(x^2-1)$ = $2x$

So,finally we have from (i),

$=2x(x^2-1)+2x(x^2+1)$

Simplifying it,we get
$4x^3$

How do you differentiate f(x)=(x^2+1)(x^2-1)f(x)=(x^2+1)(x^2-1) using the product rule?