Nigerian Scholars » Past Questions » Mathematics » Joint Admissions and Matriculation Board (JAMB) Past Question

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


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

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