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

## Dicussion (1)

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