Question

How do you differentiate `f(x)=(4x-3)/sqrt (2x^2+1)` using the quotient rule?

Answer 1

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

Explanation:

The quotient rule: if `f(x)=g(x)/(h(x)`,

`f'(x)=(g'(x)h(x)-g(x)h'(x))/(g(x))^2`

Thus,

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

Find each derivative:

`color(white)(sss)d/dx(4x-3)=4`

Use the chain rule:

`color(white)(sss)d/dxsqrt(2x^2+1)=1/2(2x^2+1)^(-1/2)d/dx(2x^2+1)`

`color(white)(sss)=>1/(2sqrt(2x^2+1)) * 4x`

`color(white)(sss)=>(2x)/sqrt(2x^2+1)`

Plug these back into the equation for `f'(x)`.

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

Multiply the numerator and denominator by `sqrt(2x^2+1)`.

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

Simplify:

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