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

Question

Isabella Ackerman · Accepted Answer

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

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)#

Luke Alvarez · Answer

undefined To differentiate $ f(x) = \frac{{4x - 3}}{{\sqrt{{2x^2 + 1}}}} $ using the quotient rule:

1. Identify $ u(x) = 4x - 3 $ and $ v(x) = \sqrt{{2x^2 + 1}} $.
2. Apply the quotient rule: $ f'(x) = \frac{{v(x) \cdot u'(x) - u(x) \cdot v'(x)}}{{[v(x)]^2}} $.
3. Find $ u'(x) $ and $ v'(x) $.
4. Substitute the values into the quotient rule formula to find $ f'(x) $.