Question

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Please evaluate the Derivative using the Limit Process. Show all workings.

Answer 1

Answer:

$\displaystyle f'(x)=-\frac{2}{x^{{}^{3}\!/\!{}_{2}}}$

Step-by-step explanation:

We have the function:

$\displaystyle f(x)=\frac{4}{\sqrt x}$

And we want to find the derivative using the limit process.

Recall that the definition of a derivative is:

$\displaystyle \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Therefore, by substitution:

$\displaystyle \lim_{h \to 0}\frac{\dfrac{4}{\sqrt{x+h}}-\dfrac{4}{\sqrt x}}{h}$

First and foremost, we can move the constant factor outside of the limit:

$\displaystyle =\lim_{h \to 0}\frac{4\left(\dfrac{1}{\sqrt{x+h}}-\dfrac{1}{\sqrt x}\right)}{h}\\ \\=4\lim_{h \to 0}\frac{\dfrac{1}{\sqrt{x+h}}-\dfrac{1}{\sqrt x}}{h}$

Next, we can multiply everything by (√(x + h)(√x) to eliminate the fractions in the denominator. Therefore:

$\displaystyle =4\lim_{h \to 0}\frac{\dfrac{1}{\sqrt{x+h}}-\dfrac{1}{\sqrt x}}{h}\left(\frac{\sqrt{x+h}\sqrt x}{\sqrt{x+h}\sqrt x}\right)$

Distribute:

$\displaystyle =4\lim_{h \to 0}\frac{\left({\sqrt{x+h}\sqrt x}\right)\dfrac{1}{\sqrt{x+h}}-(\sqrt{x+h}\sqrt x)\dfrac{1}{\sqrt x}}{h({\sqrt{x+h}\sqrt x})}$

Distribute and simplify:

$\displaystyle =4 \lim_{h\to 0}\frac{\sqrt x-\sqrt{x+h}}{h(\sqrt{x+h}\sqrt{x}) }$

Next, we can multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x + h)). Thus:

$\displaystyle = 4\lim_{h\to 0}\frac{\sqrt x-\sqrt{x+h}}{h(\sqrt{x+h}\sqrt{x}) }\left(\frac{\sqrt x +\sqrt{x+h}}{\sqrt x +\sqrt{x+h}\right)}$

Simplify:

$\displaystyle =4 \lim_{h \to 0} \frac{x-(x+h)}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}\\ \\ \\ =4 \lim_{h \to 0} \frac{x-x-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})} \\ \\ \\=4 \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Cancel like terms:

$\displaystyle =4 \lim_{h \to 0} -\frac{1}{(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Now, we can use direct substitution. Hence:

$\displaystyle \Rightarrow4 \left( -\frac{1}{(\sqrt{x+0}\sqrt x)(\sqrt x+\sqrt{x+0})}\right)$

Simplify:

$\displaystyle =4\left( -\frac{1}{(\sqrt{x}\sqrt x)(\sqrt x+\sqrt{x})}\right) \\ \\ \\ =4\left( -\frac{1}{(x)(2\sqrt{x})}\right)$

Multiply:

$\displaystyle =- \frac{4}{2x\sqrt{x}}$

Reduce and rewrite:

$\displaystyle =-\frac{2}{x(x^{{}^{1}\! / \! {}_{2} \!})}$

Simplify:

$\displaystyle =-\frac{2}{x^{{}^{3}\!/\!{}_{2}}}$

Therefore:

$\displaystyle f'(x)=-\frac{2}{x^{{}^{3}\!/\!{}_{2}}}$