Question

find dy/dx if y= (x-1)(2x+5)

Answer 1

Explanation:

for example

First expand:

(x+1)(2x-5)=

2x^2 + 2x - 5x -5 =

2x^2–3x-5

Then differentiate:

2x^2 → 4x

-3x → -3

5 → disappears

So the answer is f’(x) = 4x-3

Answer 2

$\displaystyle \bf \frac{dy}{dx} = 4x + 3$

Given :

To find :

$\displaystyle \sf \frac{dy}{dx}$

Solution :

Step 1 of 2 :

Simplify the given function

Here the given function is

$\displaystyle \sf y = (x - 1)(2x + 5)$

$\displaystyle \sf{ \implies }y = x(2x + 5) - 1(2x + 5)$

$\displaystyle \sf{ \implies }y = 2 {x}^{2} + 5x - 2x - 5$

$\displaystyle \sf{ \implies }y = 2 {x}^{2} + 3x - 5$

Step 2 of 2 :

$\displaystyle \sf Find \: \: \frac{dy}{dx}$

$\displaystyle \sf y = 2 {x}^{2} + 3x - 5$

Differentiating both sides with respect to x we get

$\displaystyle \sf \frac{dy}{dx} = \frac{d}{dx}(2 {x}^{2} + 3x - 5)$

$\displaystyle \sf{ \implies }\frac{dy}{dx} = \frac{d}{dx}(2 {x}^{2} )+ \frac{d}{dx}(3x) - \frac{d}{dx}(5)$

$\displaystyle \sf{ \implies }\frac{dy}{dx} = 2 \frac{d}{dx}( {x}^{2} )+3 \frac{d}{dx}(x) - \frac{d}{dx}(5)$

$\displaystyle \sf{ \implies }\frac{dy}{dx} = 2 \times 2 \times {x}^{2 - 1} +3 \times {x}^{1 - 1} - \frac{d}{dx}(5)\: \: \: \bigg[ \: \because \:\frac{d}{dx}( {x}^{n} ) = n {x}^{n - 1} \bigg]$

$\displaystyle \sf{ \implies }\frac{dy}{dx} = 4x +3 {x}^{0} - 0\: \: \: \bigg[ \: \because \:\frac{d}{dx}(c ) = 0 \bigg]$

$\displaystyle \sf{ \implies }\frac{dy}{dx} = 4x +3$

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