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#