Question 1

$y = \simplify[all,!noLeadingMinus]{{a}x^{n_1} + {b}x^{n_2}}$

a)

$\frac{dy}{dx} = $question.score feedback.none Expected answer:

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b)

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Advice

If $y = \mathrm{f}(x) + \mathrm{g}(x)$ then $\frac{dy}{dx}=\frac{d}{dx}\mathrm{f}(x)+\frac{d}{dx}\mathrm{g}(x)$.

In this case

$\mathrm{f}(x) = \simplify{{a}x^{n_1}}$, so $\frac{d}{dx}\mathrm{f}(x) = \simplify{{a}*{n_1}x^{n_1-1}}$
$\mathrm{g}(x) = \simplify{{b}x^{n_2}}$, so $\frac{d}{dx}\mathrm{g}(x) = \simplify{{b}*{n_2}x^{n_2-1}}$

$\frac{dy}{dx}= \simplify[all,!noLeadingMinus]{{a}*{n_1}x^{n_1-1} + {b}*{n_2}x^{n_2-1}}$

The second derivative is just differentiating again:

$\begin{align}\frac{d^2y}{dx^2}&= \frac{d}{dx}\left(\simplify[all,!noLeadingMinus]{{a}*{n_1}x^{n_1-1} + {b}*{n_2}x^{n_2-1}}\right)\\ &= \simplify[all,!noLeadingMinus]{{a}*{n_1}*{n_1-1}x^{n_1-2} + {b}*{n_2}*{n_2-1}x^{n_2-2}}\\ \end{align}$

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Question 1

Question 1

Question 1

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b)

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