Question

(x+1) (x^2+2x-3) Find the differentiation.

Answer 1

Answer:

Explanation:

we can use this formula

d/dx(uv) = u' v + v' u

Let us consider

u = x+1

v = x^2 + 2x + 3

Please find the attached file

If you want you can reduce it further

I hope this will help you

Answer 2

Answer -

Formula used -

$\frac{d}{dx} (uv) = u\frac{dv}{dx} + v \frac{du}{dx}$

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$\longrightarrow$$u = x + 1$

$\longrightarrow$$v = {x}^{2} + 2x - 3$

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$\implies$$(x + 1)\times\frac{d}{dx} ({x^{2}+2x-3}) +( {x}^{2} + 2x - 3)\times\frac{d}{dx}(x + 1)$

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$\longrightarrow$$\frac{d}{dx} ({x}^{2} + 2x - 3) = 2x + 2$

$\longrightarrow$$\frac{d}{dx}(x + 1) = 1$

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$\longrightarrow$$\frac{d}{dx} (uv) = (x + 1)(2x + 2)+({x}^{2}+2x-3) (1)$

$\longrightarrow$$2{x}^{2} + 2x + 2x + 2 + {x}^{2} + 2x - 3$

$\longrightarrow$$3 {x}^{2} + 6x - 1$

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ADDITIONAL INFORMATION -

Sum rule $\longrightarrow$

$\frac{d}{dx} (u + v) = \frac{du}{dx} + \frac{dv}{dx}$

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Difference rule $\longrightarrow$

$\frac{d}{dx} (u - v) = \frac{du}{dx} - \frac{dv}{dx}$

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Quotient rule $\longrightarrow$

$\frac{d}{dx} ( \frac{u}{v} ) = \frac{v\frac{du}{dx} - u\frac{dv}{dx} }{ {v}^{2} }$

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Chain rule $\longrightarrow$

y = f(u) and u = f(x)

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

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Thanks