# Thread: How to solve derivative with this given formula

1. ## How to solve derivative with this given formula

I'm not good at fractions and radicals and I have two questions that I can't answer. Here is the questions

a. sqrt[ (3x+1) / (2x-1)]

b. 2 / [ (6x^2+5x+1) ]^2

How can I solve this using this formula f(x) = [ f(x+h) - f(x) ] / h . I will really appreciate if you can give me the full solution + explanation (process). Thanks.

2. ## Re: How to solve derivative with this given formula

If you are using first principles to find a derivative, functions with radicals are usually simplified by rationalising the numerator.

3. ## Re: How to solve derivative with this given formula

can you show me how to solve this problem?

4. ## Re: How to solve derivative with this given formula

Here is (a):

\displaystyle \begin{align*}\dfrac{d}{dx}\left(\sqrt{\frac{3x+1} {2x-1}}\right) & = \lim_{h \to 0} \dfrac{\sqrt{\frac{3(x+h)+1}{2(x+h)-1}} - \sqrt{\frac{3x+1}{2x-1}}}{h} \\ & = \lim_{h \to 0} \dfrac{\frac{\sqrt{(3x+3h+1)(2x+2h-1)}}{2x+2h-1} - \frac{\sqrt{(3x+1)(2x-1)}}{2x-1}}{h} \\ & = \lim_{h \to 0} \dfrac{(2x-1)\sqrt{(3x+3h+1)(2x+2h-1)} - (2x+2h-1)\sqrt{(3x+1)(2x-1)}}{h(2x+2h-1)(2x-1)}\cdot \dfrac{(2x-1)\sqrt{(3x+3h+1)(2x+2h-1)} + (2x+2h-1)\sqrt{(3x+1)(2x-1)}}{(2x-1)\sqrt{(3x+3h+1)(2x+2h-1)} + (2x+2h-1)\sqrt{(3x+1)(2x-1)}} \\ & = \lim_{h \to 0} \dfrac{(2x-1)^2(3x+3h+1)(2x+2h-1)-(2x+2h-1)^2(3x+1)(2x-1)}{h(2x+2h-1)(2x-1)\left((2x-1)\sqrt{(3x+3h+1)(2x+2h-1)} + (2x+2h-1)\sqrt{(3x+1)(2x-1)}\right)} \\ & = \lim_{h \to 0} -\dfrac{5h(2h+2x-1)(2x-1)}{h(2x+2h-1)(2x-1)\left((2x-1)\sqrt{(3x+3h+1)(2x+2h-1)} + (2x+2h-1)\sqrt{(3x+1)(2x-1)}\right)} \\ & = \lim_{h \to 0} -\dfrac{5}{(2x-1)\sqrt{(3x+3h+1)(2x+2h-1)} + (2x+2h-1)\sqrt{(3x+1)(2x-1)}} \\ & = -\dfrac{5}{2(2x-1)\sqrt{(3x+1)(2x-1)}}\end{align*}

From line 1 to 2, I rationalized the denominator.
From line 2 to 3, I found a common denominator and instead of dividing the whole thing by h, I multiplied by 1/h.
On the third line, I multiplied top and bottom by the conjugate of the top.
From line 3 to 4, I used the fact that $\displaystyle (a-b)(a+b) = a^2-b^2$.
From line 4 to 5, I simplified the numerator.
From line 5 to 6, I cancelled $\displaystyle h(2x+2h-1)(2x-1)$ from both the numerator and denominator.
From line 6 to 7, I plugged in h=0 and collected like terms.