**Differentiate \((2x+5)^2(x-4)\) with respect to x.**

### Question

Differentiate \((2x+5)^2(x-4)\) with respect to x.

### Options

A) \(4(2x+5)(x-4)\)

B) \(4(2x+5)(4x-3)\)

C) \((2x+5)(2x-13)\)

D) \((2x+5)(6x-11)\)

The correct answer is D.

### Explanation:

To differentiate \((2x+5)^2(x-4),\) you first need to know that it is a product function.

Using the product rule you have \(\frac{dy}{dx}=\frac{udv}{dx}+\frac{vdu}{dx}\)

Let \((2x+5)^2\) be \(u\) and \((x-4)\) be \(v.\)

To find \(du\) we use chain rule which is \(\frac{du}{dx}=\frac{du}{dw} \times \frac{dw}{dx}\)

We say let \((2x+5)\) be \(w\)

We then have a new function \(u=w^2\)

\(\frac{du}{dw}=2w\)

\(\frac{dw}{dx}=2\)

So \(\frac{du}{dx}=2 \times 2w\) which equals \(4w\) and \(w\) was \((2x+5)\)

\(\frac{du}{dx}=4(2x+5)\)

\(\frac{dv}{dx}=1\)

Substituting everything into the product rule we have:

\(\frac{dy}{dx}=(2x+5)^2(1) + 4(2x+5)(x-4)\)

\(\frac{dy}{dx}=(4x^2 + 20x + 25) + (8x^2 - 12x - 80)\)

\(\frac{dy}{dx}=12x^2 + 8x - 55\)

\(\frac{dy}{dx}=(2x+5)(6x-11)\)

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not so clear can't it beany shorter and easier.

Great share.

To differentiate \((2x+5)^2(x-4),\) you first need to know that it is a product function.

Using the product rule you have \(\frac{dy}{dx}=\frac{udv}{dx}+\frac{vdu}{dx}\)

Let \((2x+5)^2\) be \(u\) and \((x-4)\) be \(v.\)

To find \(du\) we use chain rule which is \(\frac{du}{dx}=\frac{du}{dw} \times \frac{dw}{dx}\)

We say let \((2x+5)\) be \(w\)

We then have a new function \(u=w^2\)

\(\frac{du}{dw}=2w\)

\(\frac{dw}{dx}=2\)

So \(\frac{du}{dx}=2 \times 2w\) which equals \(4w\) and \(w\) was \((2x+5)\)

\(\frac{du}{dx}=4(2x+5)\)

\(\frac{dv}{dx}=1\)

Substituting everything into the product rule we have:

\(\frac{dy}{dx}=(2x+5)^2(1) + 4(2x+5)(x-4)\)

\(\frac{dy}{dx}=(4x^2 + 20x + 25) + (8x^2 - 12x - 80)\)

\(\frac{dy}{dx}=12x^2 + 8x - 55\)

\(\frac{dy}{dx}=(2x+5)(6x-11)\)