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

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Answer 1 · 2017-03-02T13:52:42

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

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:

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

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

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

How do you differentiate f(x)= (2x^2-5)(x+1) f(x)=(2x2−5)(x+1)f(x)= (2x^2-5)(x+1) using the product rule?