Use the rules for derivatives to find the derivatives below. y=(2x+12)(1−2x); use the Product Rule

g(x)=y2=(2x+12)(1−2x)we have to find the derivative by using the product rule:The Product Rule:ddx(f(x)×g(x))=g(x)ddxf(x)−f(x)ddxg(x)=g(x)f′(x)−f(x)g′(x)Now heref(x)=(2x+12) and g(x)=(1−2x)Therefore,g′(x)=y2′=(1−2x)ddx(2x+12)+(2x+12)ddx(1−2x)=(1−2x)×2+(2x+12)(−2)=2-4x-4x-1=1-8xHence, g′(x)=y2′=1−8x

Solved: "Use the rules for derivatives to find the..."

Braxton Pugh

Answered question

2021-09-28

Use the rules for derivatives to find the derivatives below.

y = (2 x + \frac{1}{2}) (1 - 2 x)

; use the Product Rule

Answer & Explanation

sweererlirumeX

Skilled2021-09-29Added 91 answers

$g (x) = y_{2} = (2 x + \frac{1}{2}) (1 - 2 x)$
we have to find the derivative by using the product rule:
The Product Rule:
$\frac{d}{d x} (f (x) \times g (x)) = g (x) \frac{d}{d x} f (x) - f (x) \frac{d}{d x} g (x)$
$= g (x) f^{'} (x) - f (x) g^{'} (x)$
Now here
$f (x) = (2 x + \frac{1}{2}) and g (x) = (1 - 2 x)$
Therefore,
$g^{'} (x) = y_{2}^{'} = (1 - 2 x) \frac{d}{d x} (2 x + \frac{1}{2}) + (2 x + \frac{1}{2}) \frac{d}{d x} (1 - 2 x)$
$= (1 - 2 x) \times 2 + (2 x + \frac{1}{2}) (- 2)$
=2-4x-4x-1
=1-8x
Hence, $g^{'} (x) = y_{2}^{'} = 1 - 8 x$

Jeffrey Jordon

Expert2022-02-01Added 2605 answers

Answer is given below (on video)

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