# How to get derivative of a function without using the product, quotient, or chain rule?

I have a function that I need to find a derivative of without using quotient, power, or chain rules. The function is,

$$g(x) = \frac{x^2 + 2x - 1}{\sqrt x}$$

I know how to get the answer by using the quotient rule but doing it algebraically seems to not work as well. What I'm getting so far is that I have to break it all up so it looks like this,

$$\frac{x^2}{\sqrt x} + \frac{2x}{\sqrt x} - \frac 1 {\sqrt x}$$

And after simplifying it goes to,

$$x \sqrt{x} + 2 \sqrt{x} - \frac{\sqrt x}x$$

It doesn't close to actual derivative of $$g(x)$$ though, so am I doing wrong?

• Wait a minute...what rules are left? Note: after you simplified, you had yet to find the derivative. – Andrew Chin Jun 22 '20 at 22:17
• Step 1: Prove the rules. Step 2: Use them ;) – Maximilian Janisch Jun 22 '20 at 22:34
• When you say, “without using [those] rules,” do you mean that you’re supposed to use the limit definition of the derivative? – Joe Jun 22 '20 at 22:40
• All I know is that my professor gave us that function said to compute it without any rules but instead do it algebraically. Though he did use the power rule when he tried to explain it but only at the very end. – Emerald Bay Jun 22 '20 at 22:43
• To answer your question, “am I doing wrong?”, you cannot get the derivative of a function by just using algebra to express the function in an equivalent expression. All of your algebraic manipulations above are just different ways of expressing $g(x)$, not its derivative. – Joe Jun 22 '20 at 23:07

Try the definition. Just lots of calculations! \begin{align} \frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left(\frac{(x+h)^2+2(x+h)-1}{\sqrt{x+h}}-\frac{x^2+2x-1}{\sqrt{x}}\right) \\ &= -{\frac {\sqrt {x+h}{x}^{2}-\sqrt {x}{h}^{2}-2\,{x}^{3/2}h-{x}^{5/2}+2 \,\sqrt {x+h}x-2\,\sqrt {x}h-2\,{x}^{3/2}-\sqrt {x+h}+\sqrt {x}}{ \sqrt {x+h}\sqrt {x}h}} \end{align} Rationalize the numerator: $$\frac{g(x+h)-g(x)}{h} ={\frac {x{h}^{3}+4\,{x}^{2}{h}^{2}+6\,h{x}^{3}+3\,{x}^{4}+4\,x{h}^{2}+ 12\,h{x}^{2}+8\,{x}^{3}+2\,xh+2\,{x}^{2}-1}{\sqrt {x+h}\sqrt {x} \left( {x}^{5/2}+2\,{x}^{3/2}h+2\,{x}^{3/2}+\sqrt {x}{h}^{2}+\sqrt {x +h}{x}^{2}+2\,\sqrt {x}h+2\,\sqrt {x+h}x-\sqrt {x}-\sqrt {x+h} \right) }}$$ Set $$h=0$$: $$\lim_{h\to 0}\frac{g(x+h)-g(x)}{h} = {\frac {3\,{x}^{4}+8\,{x}^{3}+2\,{x}^{2}-1}{x \left( 2\,{x}^{5/2}+4\,{ x}^{3/2}-2\,\sqrt {x} \right) }}$$ This is a perfectly good answer. If you like, you can then rationalize the denominator: $$g'(x) = {\frac {3\,{x}^{2}+2\,x+1}{2{x}^{3/2}}}$$

Probably the easiest way to take this derivative is to express $$g$$ as $$g(x)=x^{3/2}+2x^{1/2}-x^{-1/2}$$ And then you can use the power rule. Out of the forms you wrote, I'd say the first one would be easier than the other two.

• Ah, right. That makes sense. I was trying to get the derivative of sqrt(x) as 1 over 2 with sqrt(x) and it was throwing me off a bit so I left it. But I understand now. Thank you so much! – Emerald Bay Jun 22 '20 at 22:41
• @Emerald Bay, your title says "without using the product, quotient, or chain rule", but the body of your question says "without using quotient, power, or chain rules." Which is it? – Joe Jun 22 '20 at 22:46

I'm not entirely sure if this is what you're going for here but it may look like the following:

$$g(x)=\frac{x^2}{x^{\frac{1}{2}}} + \frac{2x}{x^{\frac{1}{2}}} - \frac{1}{x^{\frac{1}{2}}}$$

$$g(x)=x^{\frac{1}{2}} + 2x^{\frac{1}{2}}-x^{-\frac{1}{2}}$$

$$g'(x)=\frac{1}{2\sqrt{x}} + \sqrt{x} + \frac{1}{{2\sqrt{x^3}}}$$

I assume after you add all of these up your end result should be something along the lines of the derivative you get from the quotient rule.

A very good way to do it whan you face products or quotients is logarithmic differentiation $$g(x) = \frac{x^2 + 2x - 1}{\sqrt x}\implies \log(g(x))=\log({x^2 + 2x - 1})-\frac12 \log(x)$$

$$\frac{g'(x)}{g(x)}=\frac{2x+2}{x^2 + 2x - 1}-\frac 1{2x}$$ Simplify and use $$g'(x)=g(x)\times \frac{g'(x)}{g(x)}$$ Simplify again.