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(Note, these are not 'model' answers, but explanations of how I obtain an answer. You do not need this much explanation in your answers.)

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(i) Differentiating $\\var{a2}x^{\\var{b2}}$ gives $\\simplify{{a2}*{b2}*x^({b2}-1)}$, so we get $\\simplify{{a2}*{b2}*x^({b2}-1)*cos(x)}$ as one of the terms.

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Differentiating $\\cos(x)$ gives $-\\sin(x)$ so the other term is $\\simplify{- {a2}*x^({b2})*sin(x)}$.

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Adding these together gives the answer: $\\simplify{{a2}*{b2}*x^({b2}-1)*cos(x) - {a2}*x^({b2})*sin(x)}$.

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(ii) Differentiating $\\simplify{{a4}*x^{b4} + {c4}*x^{d4}}$ gives $\\simplify{{a4}*{b4}*x^({b4}-1) + {c4}*{d4}*x^({d4}-1)}$, so we get $\\simplify{({a4}*{b4}*x^({b4}-1) + {c4}*{d4}*x^({d4}-1))ln(x)}$ as one of the terms.

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Differentiating $\\simplify{ln(x)}$ gives $\\simplify{1/x}$ so the other term is $\\simplify{({a4}*x^{b4} + {c4}*x^{d4})/x}$, which simplifies to $\\simplify{{a4}*x^({b4}-1) + {c4}*x^({d4}-1)}$

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Adding these together gives the answer: $\\simplify{({a4}*{b4}*x^({b4}-1) + {c4}*{d4}*x^({d4}-1))ln(x) + {a4}*x^({b4}-1) + {c4}*x^({d4}-1)}$.

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(iii) Differentiating $\\simplify{ln(x)}$ gives $\\simplify{1/x}$, so we get $\\simplify{{a5}*sin(x)/x}$ as one of the terms.

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Differentiating $\\simplify{{a5}*sin(x)}$ gives $\\simplify{{a5}*cos(x)}$ so the other term is $\\simplify{{a5}*cos(x)*ln(x)}$.

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Adding these together gives the answer: $\\simplify{{a5}*sin(x)/x + {a5}*cos(x)*ln(x)}$.

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Differentiate the following with respect to $x$.

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(i) $\\var{a2}x^{\\var{b2}} \\cos(x)$ [[0]]

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(ii) $(\\simplify{{a4}*x^{b4} + {c4}*x^{d4}})\\ln(x)$ [[1]]

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(iii) $\\var{a5}\\sin(x) \\ln(x)$ [[2]]

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Differentiating three functions which require using the product rule. Non-calculator. Advice included.

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Use the product rule to differentiate.

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Questions on differentiating with the chain rule and product rule.

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