How do you use the quotient rule to differentiate #y=(x^3+x)/(4x+1) #?

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Answer 1 · 2016-02-19T15:03:13

#dy/dx=(8x^3+3x^2+1)/(4x+1)^2#

The quotient rule states that

#d/dx(f(x)/g(x))=(g(x)f'(x)-f(x)g'(x))/(g(x))^2#

When we apply this to the given function #y#, we see that

#dy/dx=((4x+1)d/dx(x^3+x)-(x^3+x)d/dx(4x+1))/(4x+1)^2#

Differentiating each of these through the power rule gives:

#dy/dx=((4x+1)(3x^2+1)-(x^3+x)(4))/(4x+1)^2#

From here, distribute and combine like terms.

#dy/dx=(12x^3+4x+3x^2+1-4x^3-4x)/(4x+1)^2#

#dy/dx=(8x^3+3x^2+1)/(4x+1)^2#