If y = 2(2x + {x})², find dy/dx.

Correct Answer is : 4(2x+x)(2+12x)4 ≤ft( 2x + {x} ) ≤ft( 2 + {1}{2{x}} )4(2x+x)(2+2x1)

4 ≤ft( 2x + {x} ) ≤ft( 2 + {1}{2{x}} )

Correct Answer is : 4(2x+x)(2+12x)4 ≤ft( 2x + {x} ) ≤ft( 2 + {1}{2{x}} )4(2x+x)(2+2x1)

Test Index

WAEC Further Mathematics 2010 Paper

Question : 12

Total: 40

y = 2(2x + \sqrt{x})^{2}

\frac{dy}{dx}

Solution:

Concept:

Use the chain rule: differentiate the outer power, then multiply by the derivative of the inner function.

Explanation:

Let

u = 2x + \sqrt{x}

Then

y = 2u^{2}

Differentiate

y

with respect to

u

\frac{dy}{du} = 4u

Differentiate

u

with respect to

x

\frac{du}{dx} = 2 + \frac{1}{2\sqrt{x}}

Apply the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

Substitute

u = 2x + \sqrt{x}

\frac{dy}{dx} = 4(2x + \sqrt{x})\left(2 + \frac{1}{2\sqrt{x}}\right)

Answer:

Option B:

4\left(2x + \sqrt{x}\right)\left(2 + \frac{1}{2\sqrt{x}}\right)

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