Given:
y = ax^2 + a
Determine:
7.3.1
\( \frac{dy}{dx} \)
7.3.2
\( \frac{dy}{da} \) - NSC Mathematics - Question 10 - 2019 - Paper 1

Question 10

Given:
y = ax^2 + a
Determine:
7.3.1
\( \frac{dy}{dx} \)
7.3.2
\( \frac{dy}{da} \)
Worked Solution & Example Answer:Given:
y = ax^2 + a
Determine:
7.3.1
\( \frac{dy}{dx} \)
7.3.2
\( \frac{dy}{da} \) - NSC Mathematics - Question 10 - 2019 - Paper 1
7.3.1 \( \frac{dy}{dx} \)

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To find ( \frac{dy}{dx} ), we need to differentiate the function ( y = ax^2 + a ) with respect to ( x ).
Using the power rule of differentiation, we get:
[
\frac{dy}{dx} = \frac{d}{dx}(ax^2) + \frac{d}{dx}(a)
]
The derivative of ( ax^2 ) is ( 2ax ) and the derivative of ( a ) (a constant) is 0. Thus:
[
\frac{dy}{dx} = 2ax + 0 = 2ax
]
7.3.2 \( \frac{dy}{da} \)

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To find ( \frac{dy}{da} ), we differentiate the function ( y = ax^2 + a ) with respect to ( a ).
Differentiating, we have:
[
\frac{dy}{da} = \frac{d}{da}(ax^2) + \frac{d}{da}(a)
]
The derivative of ( ax^2 ) with respect to ( a ) is ( x^2 ) and the derivative of ( a ) is 1:
[
\frac{dy}{da} = x^2 + 1
]
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