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This question is essentially testing your learning of how to differentiate polynomials.

a) the derivative of a constant by itself is always zero.

b) the derivative of a constant multiplied by $x$ is always just that constant. Here this means $f'(x)=\\var{a1}$.

c) the derivative of a sum is the sum of the individual terms' derivatives. This question has a sum made up of a constant and a constant multiple of $x$, so you use approach taken for a and b and then add the results. Here we have

\\[r'(t)=(\\var{a0})'+(\\var{a1}t)'=0+\\var{a1}=\\var{a1}\\]

d) the derivative of a constant multiple of a power of the variable (here $t$) is the constant multiplied by the power, then multiplied by the variable to one power fewer. That is,

\\[s'(t)=(\\var{a2})(\\var{e3})t^{\\var{e3}-1}=\\simplify{{a2}{a3}t^({e3}-1)}.\\]

e) combining the approach taken for all of the above allows you to differentiate the final polynomial. Find the derivative of each individual term, then add the results to give the derivative of the function as a whole.

\\[\\frac{\\mathrm{d}y}{\\mathrm{d}x}=\\var{a1}+(\\var{e1})(\\var{a2})x^{\\var{e1}-1}+(\\var{e2})(\\var{a3})x^{\\var{e2}-1}=\\simplify{{a1}+{e1}{a2}x^({e1}-1)+{e2}{a3}x^({e2}-1)}\\]

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If $y=\\var{a0}$, then $\\displaystyle \\frac{\\mathrm{d}y}{\\mathrm{d}x}=$

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For $f(x)=\\var{a1}x$, we can differentiate to obtain $f'(x)=$

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The function $r(t)=\\simplify{{a0}+{a1}t}$ can be differentiated to give $r'(t)=$

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The derivative of $s(t)=\\var{a2}t^\\var{e3}$ is $s'(t)=$

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The derivative of the polynomial

\\[y=\\simplify{{a0}+{a1}x+{a2}x^{e1}+{a3}x^{e2}}\\]

is $\\displaystyle \\frac{\\mathrm{d}y}{\\mathrm{d}x}$

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This question provides you with practice at developing your ability to apply rules for differentiating polynomials. You may wish to check out this video first as it will help you to attack these problems.

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This question provides practice at differentiating polynomials.

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