// Numbas version: exam_results_page_options {"name": "Power Rule, Sum or Difference (Instructional)", "metadata": {"description": "
Fairly simple questions using differentiation \"power rule\" and \"sum or difference rules\" to differentiate single term functions and polynomials.
\nSome co-efficients and indices can be negative and/or fractional.
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\nAll co-efficients and powers are integer (though some may be negative.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "You can find the derivative for powers of functions using the following rule:
\nIf \\( y=ax^n \\) then \\( \\frac{dy}{dx} = n \\times a x^{n-1} \\)
\nThe derivative of \\( f(x) + g(x) \\) is \\( \\frac{df}{dx} + \\frac{dg}{dx} \\) and the derivative of \\( f(x) - g(x) \\) is \\( \\frac{df}{dx} - \\frac{dg}{dx} \\)
\n", "advice": "We are asked to differentiate a variety of functions, each consisting of a single term.
\nWe can do this using the \"Power Rule\" for differentiation:
\nIf \\( y=ax^n \\) then \\( \\frac{dy}{dx} = n \\times a x^{n-1} \\)
\nIn plain language, \"multiply by the power, then reduce the power by one\".
\nThen:
\na)
\n\\( y= \\var{a_1} x^{\\var{n_1}} \\)
\n\\( \\frac{dy}{dx } = \\var{n_1} \\times \\var{a_1} x^{\\var{n_1} - 1} \\)
\n\\( \\frac{dy}{dx } = \\simplify{ {n_1}*{a_1}*x^{{n_1} - 1} } \\)
\n\nb)
\n\\( y= \\var{a_2} x^{\\var{n_2}} \\)
\n\\( \\frac{dy}{dx } = \\var{n_2} \\times \\var{a_2} x^{\\var{n_2} - 1} \\)
\n\\( \\frac{dy}{dx } = \\simplify{ {n_2}*{a_2}*x^{{n_2} - 1} } \\)
\nc)
\n\\( y= \\var{a_3} x^{\\var{n_3}} \\)
\n\\( \\frac{dy}{dx } = \\var{n_3} \\times \\var{a_3} x^{\\var{n_3} - 1} \\)
\n\\( \\frac{dy}{dx } = \\simplify{ {n_3}*{a_3}*x^{{n_3} - 1} } \\)
\n\n
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Co-efficient for Q1
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\n\\( \\frac{dy}{dx } = \\)[[0]]
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\n\\( \\frac{dy}{dx } = \\) [[0]]
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\n\\( \\frac{dy}{dx } = \\) [[0]]
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\nSome co-efficients and powers are non-integer and some may be negative.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "You can find the derivative for powers of functions using the following rule:
\nIf \\( y=ax^n \\) then \\( \\frac{dy}{dx} = n \\times a x^{n-1} \\)
\nThe derivative of \\( f(x) + g(x) \\) is \\( \\frac{df}{dx} + \\frac{dg}{dx} \\) and the derivative of \\( f(x) - g(x) \\) is \\( \\frac{df}{dx} - \\frac{dg}{dx} \\)
\n", "advice": "We are asked to differentiate a variety of functions, each consisting of a single term.
\nWe can do this using the \"Power Rule\" for differentiation:
\nIf \\( y=ax^n \\) then \\( \\frac{dy}{dx} = n \\times a x^{n-1} \\)
\nIn plain language, \"multiply by the power, then reduce the power by one\".
\nThen:
\na)
\n\\( y= \\var{a_1} x^{\\var{n_1}} \\)
\n\\( \\frac{dy}{dx } = \\var{n_1} \\times \\var{a_1} x^{\\var{n_1} - 1} \\)
\n\\( \\frac{dy}{dx } = \\simplify{ {n_1}*{a_1}*x^{{n_1} - 1} } \\)
\n\nb)
\n\\( y= \\var{a_2} x^{\\var[fractionNumbers]{n_2}} \\)
\n\\( \\frac{dy}{dx } = \\var{n_2} \\times \\var{a_2} x^{\\var[fractionNumbers]{n_2} - 1} \\)
\n\\( \\frac{dy}{dx } = \\simplify{ {n_2}*{a_2}*x^{{n_2} - 1} } \\)
\nc)
\n\\( y= \\var{a_3} x^{\\var{n_3}} \\)
\n\\( \\frac{dy}{dx } = \\var{n_3} \\times \\var{a_3} x^{\\var{n_3} - 1} \\)
\n\\( \\frac{dy}{dx } = \\simplify{ {n_3}*{a_3}*x^{{n_3} - 1} } \\)
\n\n
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Co-efficient for Q1
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\n\\( \\frac{dy}{dx } = \\)[[0]]
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\n\\( \\frac{dy}{dx } = \\) [[0]]
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\nSome co-efficients and powers are non-integer and some may be negative.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "You can find the derivative for powers of functions using the following rule:
\nIf \\( y=ax^n \\) then \\( \\frac{dy}{dx} = n \\times a x^{n-1} \\)
\nThe derivative of \\( f(x) + g(x) \\) is \\( \\frac{df}{dx} + \\frac{dg}{dx} \\) and the derivative of \\( f(x) - g(x) \\) is \\( \\frac{df}{dx} - \\frac{dg}{dx} \\)
\n", "advice": "We are asked to differentiate a variety of functions, each consisting of a single term.
\nWe can do this using the \"Power Rule\" for differentiation:
\nIf \\( y=ax^n \\) then \\( \\frac{dy}{dx} = n \\times a x^{n-1} \\)
\nIn plain language, \"multiply by the power, then reduce the power by one\".
\nThe \"Sum or Difference Rules\" also tell us that, as long as the terms are either added or subtracted, we can differentiate the function term by term.
\nThen:
\na)
\n\\( y= \\var{a1} x^{\\var{n1}} + \\var{a2} x^{\\var{n2}} +\\var{a3} \\)
\n\\( \\frac{dy}{dx } =\\var{n1} \\times \\var{a1} x^{\\var{n1} - 1} + \\var{n2} \\times \\var{a2} x^{\\var{n2}-1} \\)
\nThe constant term ( \\( \\var{a3} \\) ) can be seen as \\( \\var{a3}x^0 \\) so will differentiate to \\( 0 \\times \\var{a3} x^{0-1} \\) which, of course equals zero.
\n\\( \\frac{dy}{dx } = \\simplify{ {n1}*{a1}*x^({n1}-1) + {n2}*{a2}*x^({n2}-1) } \\)
\n\nb)
\n\\( \\simplify{y= {b1} x^{{m1}} + {b2} x^{{m2}}+{b3}x^{{m3}}+{b4} } \\)
\n\\( \\frac{dy}{dx } = \\var{m1} \\times \\var{b1} x^{\\var{m1} - 1} +\\var{m2} \\times \\var{b2} x^{\\var{m2} - 1}+\\var{m3} \\times \\var{b3} x^{\\var{m3} - 1}\\)
\nThe constant term ( \\( \\var{b4} \\) ) can be seen as \\( \\var{b4}x^0 \\) so will differentiate to \\( 0 \\times \\var{b4} x^{0-1} \\) which, of course equals zero.
\n\\( \\frac{dy}{dx } =\\simplify{ {m1}*{b1}*x^({m1}-1) + {m2}*{b2}*x^({m2}-1)+ {m3}*{b3}*x^({m3}-1) } \\)
\n\nc)
\n\\( \\simplify{y= {c1} x^{{p1}} + {c2} x^{{p2}} } \\)
\nrem \\( \\frac{dy}{dx } = \\var{p1} \\times \\var{c1} x^{\\var{p1} - 1} +\\var[fractionNumbers]{p2} \\times \\var{c2} x^{\\var[fractionNumbers]{p2} - 1} \\)
\n\\( \\frac{dy}{dx } = \\simplify{{p1}*{c1}*x^{{p1}-1} + {p2}*{c2}*x^{{p2}-1} } \\)
\n\n
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\n\\( \\frac{dy}{dx } = \\) [[0]]
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\nSome co-efficients and powers are non-integer and some may be negative.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "You can find the derivative for powers of functions using the following rule:
\nIf \\( y=ax^n \\) then \\( \\frac{dy}{dx} = n \\times a x^{n-1} \\)
\nThe derivative of \\( f(x) + g(x) \\) is \\( \\frac{df}{dx} + \\frac{dg}{dx} \\) and the derivative of \\( f(x) - g(x) \\) is \\( \\frac{df}{dx} - \\frac{dg}{dx} \\)
\n", "advice": "We are asked to differentiate a variety of functions, each consisting of a single term.
\nWe can do this using the \"Power Rule\" for differentiation:
\nIf \\( y=ax^n \\) then \\( \\frac{dy}{dx} = n \\times a x^{n-1} \\)
\nIn plain language, \"multiply by the power, then reduce the power by one\".
\nThe \"Sum or Difference Rules\" also tell us that, as long as the terms are either added or subtracted, we can differentiate the function term by term.
\nThen:
\na)
\n\\( y= \\var{a1} x^{\\var{n1}} + \\var{a2} x^{\\var{n2}} +\\var{a3} \\)
\n\\( \\frac{dy}{dx } =\\var{n1} \\times \\var{a1} x^{\\var{n1} - 1} + \\var{n2} \\times \\var{a2} x^{\\var{n2}-1} \\)
\nThe constant term ( \\( \\var{a3} \\) ) can be seen as \\( \\var{a3}x^0 \\) so will differentiate to \\( 0 \\times \\var{a3} x^{0-1} \\) which, of course equals zero.
\n\\( \\frac{dy}{dx } = \\simplify{ {n1}*{a1}*x^({n1}-1) + {n2}*{a2}*x^({n2}-1) } \\)
\n\nb)
\n\\( \\simplify{y= {b1} x^{{m1}} + {b2} x^{{m2}}+{b3}x^{{m3}}+{b4} } \\)
\n\\( \\frac{dy}{dx } = \\var{m1} \\times \\var{b1} x^{\\var{m1} - 1} +\\var{m2} \\times \\var{b2} x^{\\var{m2} - 1}+\\var{m3} \\times \\var{b3} x^{\\var{m3} - 1}\\)
\nThe constant term ( \\( \\var{b4} \\) ) can be seen as \\( \\var{b4}x^0 \\) so will differentiate to \\( 0 \\times \\var{b4} x^{0-1} \\) which, of course equals zero.
\n\\( \\frac{dy}{dx } =\\simplify{ {m1}*{b1}*x^({m1}-1) + {m2}*{b2}*x^({m2}-1)+ {m3}*{b3}*x^({m3}-1) } \\)
\n\nc)
\n\\( \\simplify{y= {c1} x^{{p1}} + {c2} x^{{p2}} } \\)
\nrem \\( \\frac{dy}{dx } = \\var{p1} \\times \\var{c1} x^{\\var{p1} - 1} +\\var[fractionNumbers]{p2} \\times \\var{c2} x^{\\var[fractionNumbers]{p2} - 1} \\)
\n\\( \\frac{dy}{dx } = \\simplify{{p1}*{c1}*x^{{p1}-1} + {p2}*{c2}*x^{{p2}-1} } \\)
\n\n
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