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

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)
$$(2x+5)(6x-11)$$

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:
$$\frac{dy}{dx}=(2x+5)^2(1) + 4(2x+5)(x-4)$$
$$\frac{dy}{dx}=(4x^2 + 20x + 25) + (8x^2 - 12x - 80)$$
$$\frac{dy}{dx}=12x^2 + 8x - 55$$
$$\frac{dy}{dx}=(2x+5)(6x-11)$$

Discussion (13)

• Marvellous

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

• Chinonso

Please send me the solution of this mathematics

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

1. Really

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

• Not to very clear ☺️☺️

• Samuel

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.

• Espenschied

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:
$$\frac{dy}{dx}=(2x+5)^2(1) + 4(2x+5)(x-4)$$
$$\frac{dy}{dx}=(4x^2 + 20x + 25) + (8x^2 - 12x - 80)$$
$$\frac{dy}{dx}=12x^2 + 8x - 55$$
$$\frac{dy}{dx}=(2x+5)(6x-11)$$

1. Gabriel

Please where did you get the 20x

2. It was gotten from expanding $$(2x + 5)^2$$.
$$(2x + 5)(2x + 5) = 4x^2 + 10x + 10x + 25 = 4x^2 + 20x + 25$$