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

### Question

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

### Options

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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First u need to understand the question is aboult product function.. Lets start solving..

all u just have to do is

add the first product which is (2x plus 5)

to the second product which is (x minus 4)

...giving you (x minus 11) answer (D)....

No need for the long explannation

Please send me the solution of this mathematics

⚖️ not so balance to understanding due to the dw in the chain rule

Really

I understand this very much...... If you don't understand DM me

Not to very clear ☺️☺️

Maths

Pls not clear enough for me to understand pls can you explain to me like what a secondary school 1 would understand

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:

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

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

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

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

Please where did you get the 20x

It was gotten from expanding \( (2x + 5)^2 \).

\( (2x + 5)(2x + 5)

= 4x^2 + 10x + 10x + 25

= 4x^2 + 20x + 25 \)