How do you differentiate #f(x)= (2x^2-5)(x+1) # using the product rule?

1 Answer
Mar 2, 2017

Answer:

#f'(x)=6x^2+4x-5#

Explanation:

The product rule is #k'(x)=f(x)g'(x)+g(x)f'(x)#

Written informally, this is:

(derivative of product)=(first)x(derivative of second)+(second)x(derivative of the first).

In this case the 'first' is #(2x^2-5)# and the 'second' is #(x+1)#.

The derivative of the 'first' is calculated as follows:
#d/dx(2x^2-5)=d/dx2x^2+d/dx-5#

#=2d/dxx^2-d/dx5#

#=2*2x-0#

#=4x#

The derivative of the 'second' is calculated:

#d/dx(x+1)=d/dxx+d/dx1#

#=1+0=1#

Using the product rule (and the derivatives worked out above) gives:

#f'(x)=(2x^2-5)(1)+(x+1)(4x)#

#=2x^2-5+4x^2+4x#

#=6x^2+4x-5#