Question

hi! trying to find a difference quotient, but how do I approach this when there are fractions in the numerator of a bigger fraction?

Answer 1

`DQ=((1)/(2x+2h)-(1)/(2x))/(h)`

start by solving the subtraction of fractions in the numerator using the subtraction of fractions

`(a)/(b)-(c)/(d)=(a\cdot d-b\cdot c)/(bd)`

Then, apply to the fractions in the numerator

`(1)/(2x+2h)-(1)/(2x)=(2x-(2x+2h))/(2x\cdot(2x+2h))`

Simplify the expression at the numerator

`(2h)/(4x^2+4xh)`

`DQ=((2x-2x-2h)/(4x^2+4hx))/(h)`

simplify

`DQ=(-(h)/(2x^2+2xh))/(h)`

then apply the division of fractions

`\begin{gathered} DQ=(-h)/(h\cdot(2x^2+2xh)) \\ DQ=-(1)/(2x^2+2xh) \end{gathered}`

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