How do you use the quotient rule to differentiate #x/ sqrt (x^2 + 1)#?

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Answer 1 · 2016-10-03T11:58:23

#\frac{1}{(x^2+1)^{3/2}}#

The quotient rule states that

#d/dx (f/g) = \frac{f'g - fg'}{g^2}#

Let's compute everything's required, and then we'll just substitute our results in this formula: we have

#f(x) = x \implies f'(x) = 1#

#g(x) = sqrt(x^2+1) \implies g'(x) = (x)/(sqrt(x^2+1))#

Also, #g^2(x) = x^2+1#

Thus,

#\frac{f'g - fg'}{g^2} = \frac{1*sqrt(x^2+1) - x*(x)/(sqrt(x^2+1))}{x^2+1}#

We can write this last result as

#\frac{sqrt(x^2+1) -(x^2)/(sqrt(x^2+1))}{x^2+1}#

And then we can rearrange the numerator as

#\frac{\frac{x^2+1-x^2}{sqrt(x^2+1)}}{x^2+1}#

which equals

#\frac{1}{sqrt(x^2+1)(x^2+1)}=\frac{1}{(x^2+1)^{3/2}}#