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I get that $$\frac{d}{dx}\left(7\times\dfrac{1}{\sqrt{x}}\right)=\frac{d}{dx}(7x^{.5})=\dfrac{7}{2}x^{-.5}$$ is the derivative, but I can't ever use $\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.

If someone or anyone could go step by step and do the problem, I would be eternally grateful.

  • 0
    For your information, note that: $$\frac{d}{dx}\left(7\times \frac{1}{\sqrt{x}}\right)=\frac{d}{dx}\left(7x^{\color{red}{-0.5}}\right)$$ So that is why the answers below are giving different answers to the one you've evaluated without first principles. – 2017-02-11

5 Answers 5

4

$$\begin{align}f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\to0}\frac{\frac{7}{\sqrt{x+h}}-\frac{7}{\sqrt x}}{h}\\ &=\lim_{h\to0}\frac{7\sqrt x-7\sqrt{x+h}}{h\sqrt{x(x+h)}}\\ &=7\lim_{h\to0}\frac{\sqrt x-\sqrt{x+h}}{h\sqrt{x(x+h)}}\cdot\frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}}\\ &=7\lim_{h\to0}\frac{-h}{h\sqrt{x(x+h)}(\sqrt{x}+\sqrt{x+h})}\\ &=7\lim_{h\to0}\frac{-1}{\sqrt{x(x+h)}(\sqrt{x}+\sqrt{x+h})}\\ &=-\frac{7}{2x^\frac{3}{2}}\end{align}$$

2

$$\begin{align*}f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &= \lim_{h\to 0} \frac7h\cdot \left( \frac1{\sqrt{x+h}} -\frac1{\sqrt{x}} \right)\\ &= \lim_{h\to 0} \frac7h\cdot\left(\frac{\sqrt x-\sqrt{x+h}}{\sqrt{x(x+h)}} \right)\\ &= \lim_{h\to 0} \frac7h\cdot\left(\frac{\sqrt x-\sqrt{x+h}}{\sqrt{x(x+h)}} \cdot \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}} \right)\\ &= \lim_{h\to 0} \frac7h\cdot\left(\frac{x-(x+h)}{\sqrt{x(x+h)}\cdot\left(\sqrt{x+h}+\sqrt x\right)}\right)\\ &=\lim_{h\to 0}\frac{-7}{\sqrt{x(x+h)}\cdot\left(\sqrt{x+h}+\sqrt x\right)}\\ &=\frac{-7}{x\cdot 2\sqrt x} \end{align*}$$

1

\begin{align*} \dfrac{\frac{7}{\sqrt{x+h}} - \frac{7}{\sqrt{x}}}{h} &= 7\left(\dfrac{\sqrt x - \sqrt{x+h}}{h\sqrt{x(x+h)}}\right) \\ &= 7\left(\dfrac{\sqrt x - \sqrt{x+h}}{h\sqrt{x(x+h)}}\right)\left(\dfrac{\sqrt x + \sqrt{x+h}}{\sqrt x + \sqrt{x+h}}\right) \\ &= 7\left(\dfrac{x-(x+h)}{h\sqrt{x(x+h)}}\right)\left(\dfrac{1}{\sqrt x + \sqrt{x+h}}\right) \\ &= 7\left(\dfrac{-h}{h\sqrt{x(x+h)}}\right)\left(\dfrac{1}{\sqrt x + \sqrt{x+h}}\right) \\ &= 7\left(\dfrac{-1}{\sqrt{x(x+h)}}\right)\left(\dfrac{1}{\sqrt x + \sqrt{x+h}}\right) \\ \end{align*}

Therefore,

$$\lim_{h \to 0}\dfrac{\frac{7}{\sqrt{x+h}} - \frac{7}{\sqrt{x}}}{h} = \lim_{h \to 0} 7\left(\dfrac{-1}{\sqrt{x(x+h)}}\right)\left(\dfrac{1}{\sqrt x + \sqrt{x+h}}\right) = \frac{-7}{2|x|\sqrt x} = \frac{-7}{2x^{3/2}}$$

0

$$f(x)=\dfrac{7}{\sqrt{x}}\\f'(x)=\lim\limits_{h \to0}\dfrac{f(x+h)-f(x)}{h}\\f'(x)=\lim\limits_{h \to0}\dfrac{\frac{7}{\sqrt{x+h}}-\frac{7}{\sqrt{x}}}{h}\\f'(x)=\lim\limits_{h \to0}\dfrac{\bigg(\frac{7}{\sqrt{x+h}}-\frac{7}{\sqrt{x}}\bigg)\bigg(\frac{7}{\sqrt{x+h}}+\frac{7}{\sqrt{x}}\bigg)}{h\bigg(\frac{7}{\sqrt{x+h}}+\frac{7}{\sqrt{x}}\bigg)}\\f'(x)=\lim\limits_{h \to0}\dfrac{\frac{49}{x+h}-\frac{7}{\sqrt{x}}}{h\bigg(\frac{7}{\sqrt{x+h}}+\frac{7}{\sqrt{x}}\bigg)}\\f'(x)=\lim\limits_{h \to0}\dfrac{49x-49x-49h}{h\bigg(\frac{7}{\sqrt{x+h}}+\frac{7}{\sqrt{x}}\bigg)(x(x+h))}\\f'(x)=\lim\limits_{h \to0}\dfrac{-49}{\bigg(\frac{7}{\sqrt{x+h}}+\frac{7}{\sqrt{x}}\bigg)(x(x+h))}\\f'(x)=\dfrac{-49}{(\frac{14}{\sqrt{x}})x^2}\\f'(x)=\dfrac{-7}{2x^{\frac{3}{2}}}\\f'(x)=-\dfrac{7}{2}x^{-\frac{3}{2}}$$

-1

Hint:

Use binomial series expansion of $(x+h)^{\frac{-1}{2}}$

  • 0
    I promise you the OP does not have series methods at their disposal yet. – 2017-02-11
  • 0
    @AlfredYerger Ok i got it.. – 2017-02-11