How to ifferentiate using first principles for function $$f(x)= x\sqrt{x}$$ Can anyone help i am really stuck, I'm not entirely sure how to set out the question and also how to answer it.
Differentiate using first principles for $f(x)= x\sqrt{ x}$?
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$\begingroup$
calculus
derivatives
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0Can you type it out on the box below? – 2017-02-04
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0It seems to be asking you to use the difference quotient limit definition of the derivative. Are you familiar with that definition? If so, what difficulty did you have in proceeding? – 2017-02-04
1 Answers
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If you are using the definition
$$f^\prime(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
then let $f(x)=x\sqrt{x}$. You will need to use "complete the square."
\begin{eqnarray} f^\prime(x)&=&\lim_{h\to0}\frac{(x+h)\sqrt{x+h}-x\sqrt{x}}{h}\\ &=&\lim_{h\to0}\frac{(x+h)\sqrt{x+h}-x\sqrt{x}}{h}\cdot\frac{(x+h)\sqrt{x+h}+x\sqrt{x}}{(x+h)\sqrt{x+h}+x\sqrt{x}}\\ &=&\lim_{h\to0}\dfrac{(x+h)^3-x^3}{h((x+h)\sqrt{x+h}+x\sqrt{x})}\\ &=&\lim_{h\to0}\dfrac{(3x^2+3xh+h^2)h}{h((x+h)\sqrt{x+h}+x\sqrt{x})}\\ &=&\lim_{h\to0}\dfrac{3x^2+3xh+h^2}{(x+h)\sqrt{x+h}+x\sqrt{x}}\\ &=&\frac{3x^2}{2x\sqrt{x}}\\ &=&\frac{3\sqrt{x}}{2} \end{eqnarray}
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2Could the questioner comment to indicate whether or not this is what was meant by "first principles" or whether you had something else in mind? – 2017-02-04