Mathematics (2unit) Differentiation Questions Help D: [3 Questions] (1 Viewer)

MysteriousZ · Mar 13, 2017

Im not sure how to solve these correctly D:

i have to solve using first principles method

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Mahan1 · Mar 19, 2017

MysteriousZ said:
Im not sure how to solve these correctly D:

i have to solve using first principles method

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$\textrm{a)} \ \frac{d}{dx}(x^{\frac{5}{3}}) = \lim_{h\to 0} \frac{(x+h)^{\frac{5}{3}} - x^{\frac{5}{3}} }{h}$

$= \lim_{h\to 0} \frac{(x+h)(x+h)^{\frac{2}{3}} - x x^{\frac{2}{3}}}{h} = \lim_{h \to 0} x (\frac{(x+h) ^{\frac{2}{3}} - x^{\frac{2}{3}} }{h}) + \frac{h (x+h) ^{\frac{2}{3}} }{h}$

$= x^{\frac{2}{3}} + \lim_{h \to 0} x (\frac{(x+h) ^{\frac{2}{3}} - x^{\frac{2}{3}} }{h}) = x^{\frac{2}{3}} + \lim_{h \to 0}x(\frac{(x+h)^{2} - x^{2}}{h((x+h)^{\frac{4}{3}} + (x (x+h)) ^{\frac{2}{3}} + x^{\frac{4}{3}})})$

$= x^{\frac{2}{3}} + \lim_{h \to 0} \frac{(x+h)^{2} - x^{2}}{h} \ \lim_{h \to 0 } \frac{1}{((x+h)^{\frac{4}{3}} + (x (x+h)) ^{\frac{2}{3}} + x^{\frac{4}{3}})}$

$= x^{\frac{2}{3}} + 2x \frac{x}{3 x^{\frac{4}{3}}} = \frac{5}{3} x^{\frac{2}{3}}$

$\textrm{I used the identity} \ a^{3} - b^{3} = (a-b) (a^{2} + ab + b^{2} ) \ \textrm{and } \ \lim_{h \to 0} \frac{(x+h)^{2} -x ^{2}}{h} = 2x$

$\textrm{b)} \lim_{h \to 0} \frac{\frac{1}{(x+h)^{\frac{3}{2}}} - \frac{1}{x^{\frac{3}{2}}} }{h} = \lim_{h \to 0} -\frac{ (x+h)^{\frac{3}{2}} - x^{\frac{3}{2}}}{h(x(x+h))^{\frac{3}{2}}}$

$= -\lim_{h \to 0} \frac{(x+h)^{3} - x^{3}}{h(x(x+h))^{\frac{3}{2}}((x+h)^{3} + x^{3})}= - \lim_{h \to 0}\frac{(x+h)^{3} - x^{3}}{h} \ \lim_{h \to 0}\frac{1}{(x(x+h))^{\frac{3}{2}}((x+h)^{\frac{3}{2}} + x^{\frac{3}{2}})} = -3x^{2} \frac{1}{2x^{3}x^{\frac{3}{2}}} = -\frac{3}{2}x^{\frac{-5}{2}}$

$\textrm{c)} \lim_{h \to 0} \frac{\frac{5(x+h)^{3} + 3(x+h)^{2} - 4}{x+h} - \frac{5x^{3} + 3x^{2} - 4}{x}}{h} = \lim_{h \to 0} \frac{x(5(x+h)^{3} + 3(x+h)^{2} -4) - (x+h)(5x^{3} + 3x^{2} -4)}{h(x(x+h))}$

$\lim_{h\to 0 } \frac{5x^{4} + 15x^{3}h + 15 x^{2} h^{2} + 5x h^{3} + 3x^{3} + 6x^{2}h + 3h^{2} x -4x - 5x^{4} -3x^{3} +4x -5x^{3}h - 3x^{2}h + 4h}{h(x(x+h))}$

$=\lim{h \to 0} \frac{15x^{2}h^{2} + 5xh^{3} + 3xh^{2}10x^{3}h + 3x^{2}h+4}{h (x(x+h))} = \frac{10x^{3}+ 3x^{2} + 4}{x^{2}}$

Basically, the technique is when you see
$x^{\frac{m}{n} - y ^{\frac{m}{n}}}$ then use $a^{m} - b^{m}$ expansion, to find the limit.

Mathematics (2unit) Differentiation Questions Help D: [3 Questions] (1 Viewer)

MysteriousZ

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Mahan1

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