Page:Calculus Made Easy.pdf/63

(6) Differentiate ${\displaystyle y=0.5x^{3}(x-3)}$

${\displaystyle {\frac {dy}{dx}}}$	${\displaystyle =0.5\left[x^{3}{\frac {d(x-3)}{dx}}+(x-3){\frac {d(x^{3})}{dx}}\right]}$
	${\displaystyle =0.5\left[x^{3}+(x-3)\times 3x^{2}\right]=2x^{3}-4.5x^{2}}$.

Same remarks as for preceding example.

(7) Differentiate ${\displaystyle w=\left(\theta +{\dfrac {1}{\theta }}\right)\left({\sqrt {\theta }}+{\dfrac {1}{\sqrt {\theta }}}\right)}$

This may be written

${\displaystyle {\frac {dw}{d\theta }}}$	${\displaystyle =(\theta +\theta ^{-1}){\frac {d(\theta ^{\frac {1}{2}}+\theta ^{-{\frac {1}{2}}})}{d\theta }}+(\theta ^{\frac {1}{2}}+\theta ^{-{\frac {1}{2}}}){\frac {d(\theta +\theta ^{-1})}{d\theta }}}$
	${\displaystyle =(\theta +\theta ^{-1})({\tfrac {1}{2}}\theta ^{-{\frac {1}{2}}}-{\tfrac {1}{2}}\theta ^{-{\frac {3}{2}}})+(\theta ^{\frac {1}{2}}+\theta ^{-{\frac {1}{2}}})(1-\theta ^{-2})}$
	${\displaystyle ={\tfrac {1}{2}}(\theta ^{\frac {1}{2}}+\theta ^{-{\frac {3}{2}}}-\theta ^{-{\frac {1}{2}}}-\theta ^{-{\frac {5}{2}}})+(\theta ^{\frac {1}{2}}+\theta ^{-{\frac {1}{2}}}-\theta ^{-{\frac {3}{2}}}-\theta ^{-{\frac {5}{2}}})}$
	${\displaystyle ={\tfrac {3}{2}}\left({\sqrt {\theta }}-{\frac {1}{\sqrt {\theta ^{5}}}}\right)+{\tfrac {1}{2}}\left({\frac {1}{\sqrt {\theta }}}-{\frac {1}{\sqrt {\theta ^{3}}}}\right)}$.

This, again, could be obtained more simply by multiplying the two factors first, and differentiating afterwards. This is not, however, always possible; see, for instance, p. 173, example 8, in which the rule for differentiating a product must be used.

(8) Differentiate ${\displaystyle y={\frac {a}{1+a{\sqrt {x}}+a^{2}x}}}$.

${\displaystyle {\frac {dy}{dx}}}$	${\displaystyle ={\frac {(1+ax^{\frac {1}{2}}+a^{2}x)\times 0-a{\frac {d(1+ax^{\frac {1}{2}}+a^{2}x)}{dx}}}{(1+a{\sqrt {x}}+a^{2}x)^{2}}}}$
	${\displaystyle =-{\frac {a({\frac {1}{2}}ax^{-{\frac {1}{2}}}+a^{2})}{(1+ax^{\frac {1}{2}}+a^{2}x)^{2}}}}$.