How do you find the first and second derivatives of #y=(x^3+x)/(4x+1)# using the quotient rule?

Question

Aurora Alvarado · Accepted Answer

#dy/dx f(x)/g(x) = (g(x)f'(x)-f(x)g'(x))/g(x)^2# so you must find the derivative of the numerator and the denominator and plug them into this "formula".

The quotient rule goes as follows: #dy/dx f(x)/g(x) = (g(x)f'(x)-f(x)g'(x))/g(x)^2# An easy way to remember this is the mnemonic "low d hi - hi d low over low squared." D meaning derivative. So the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator divided by the denominator squared. So, if we write down the function next to it's derivatives, it would look like this: #(x^3+x)/(4x+1)# derivatives: #(3x^2+1)/4# We find the derivatives of the numerator and denominator using the power rule. Now that we have that, we plug in each value into the quotient rule. #((4x+1)(3x^2+1)-4(x^3+x))/(4x+1)^2# Now this expression above is your first derivative. Now in order to find the second derivative, you must take the derivative of the first derivative, but I'm going to clean it up first. #((12x^3+3x^2+4x+1)-(4x^3+4x))/(16x^2+8x+1)# Which simplifies to (Going to go ahead and put the derivatives here): #(8x^3+3x^2+1)/(16x^2+8x+1)# Derivatives: #(24x^2+6x)/(32x+8)# Now we must find the derivative of this, having fun yet? #("low"*d "hi" - "hi"*d"low")/"low"^2# #((16x^2+8x+1)(24x^2+6x)-(8x^3+3x^2+1)(32x+8))/(16x^2+8x+1)^2# Now this lovely function is your second derivative. You can clean it up if you wish, but most teachers will be okay if you leave it here, you may also factor a bit if you would like.

Samantha Adkison · Answer

undefined To find the first derivative of $ y = \frac{x^3 + x}{4x + 1} $ using the quotient rule:

1. Let $ u = x^3 + x $ and $ v = 4x + 1 $.
2. Use the quotient rule formula:
   $$ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2} $$
3. Find $ u' $ and $ v' $ (the derivatives of $ u $ and $ v $ with respect to $ x $):
   $$ u' = 3x^2 + 1 $$
   $$ v' = 4 $$
4. Plug these values into the quotient rule formula:
   $$ \frac{d}{dx} \left( \frac{x^3 + x}{4x + 1} \right) = \frac{(4x + 1)(3x^2 + 1) - (x^3 + x)(4)}{(4x + 1)^2} $$

To find the second derivative, take the derivative of the first derivative obtained above.