Question

How do you use the quotien rule to differentiate `(3x^2+4) / sqrt(1+x^2)`?

Answer 1

`f'(x)=(6xsqrt(1+x^2)-((3x^3+4x)/sqrt(1+x^2)))/(1+x^2)`

Explanation:

Quotient Rule

`f'(x)=(vu'-uv')/v^2`

`u=3x^2+4`

`u'=6x`

`v=sqrt(1+x^2)=(1+x^2)^(1/2)`

`v'=(1/cancel2)(1+x^2)^(-1/2)*cancel2x=x/sqrt(1+x^2)`

`f'(x)=((1+x^2)^(1/2)*6x-(3x^2+4)(x/sqrt(1+x^2)))/((1+x^2)^(1/2))^2`

Simplify

`f'(x)=(6xsqrt(1+x^2)-((x(3x^2+4))/sqrt(1+x^2)))/(1+x^2)`

`f'(x)=(6xsqrt(1+x^2)-((3x^3+4x)/sqrt(1+x^2)))/(1+x^2)`

Common Denominator

`f'(x)=(6xsqrt(1+x^2)*((sqrt(1+x^2))/(sqrt(1+x^2)))-((3x^3+4x)/sqrt(1+x^2)))/(1+x^2)`

Simplify

`f'(x)=(((6x(1+x^2))/(sqrt(1+x^2)))-((3x^3+4x)/sqrt(1+x^2)))/(1+x^2)`

Distribute

`f'(x)=((6x+6x^3)/(sqrt(1+x^2))-((3x^3+4x)/sqrt(1+x^2)))/(1+x^2)`

Numerator simplified

`f'(x)=((6x+6x^3-3x^3-4x)/sqrt(1+x^2))/(1+x^2)`

`f'(x)=((2x+3x^3)/sqrt(1+x^2))/(1+x^2)`

Multiply by the reciprocal

`f'(x)=(2x+3x^3)/sqrt(1+x^2)*1/(1+x^2)`

`f'(x)=(2x+3x^3)/(sqrt(1+x^2)*(1+x^2))`

Simplify

`f'(x)=(2x+3x^3)/((1+x^2)^(1/2)*(1+x^2)^(2/2))`

Simplify the denominator

`f'(x)=(2x+3x^3)/((1+x^2)^(3/2))`

Factor out an x from the numerator

`f'(x)=(x(2+3x^2))/((1+x^2)^(3/2))`

Watch these examples of the quotient rule.