How do you use the quotient rule to differentiate #y=(2x^4-3x)/(4x-1)#?

Question

Paisley Johnson · Accepted Answer

#f'(x)=(24x^4-8x^3+3)/(4x-1)^2#

The Quotient rle is given by #(u/v)'=(u'v-uv')/v^2# let us denote by #u=2x^4-3x#

and

#v=4x-1#

so we get #u'=8x^3-3# and #v'=4#

now we Can build the derivative: #f'(x)=((8x^3-3)(4x-1)-(2x^4-3x)*4)/(4x-1)^2#

multiplying out the numerator and collecting like Terms we get

#f'(x)=(24x^4-8x^3+3)/(4x-1)^2#

Luke Alvarez · Answer

#(24x^4-8x^3+4)/(4x-1)^2#

If we have a quotient of functions #f(x)# and #g(x)#, we can find the derivative with the Quotient Rule

#(f'(x)g(x)-f(x)g'(x))/(g(x))^2#

In our scenario, we basically have

#f(x)=2x^4-3x=>f'(x)=8x^3-3# and

#g(x)=4x-1=>g'(x)=4#

By entering our expressions into the Quotient Rule, we can obtain

#((8x^3-3)(4x-1)-4(2x^4-3x))/(4x-1)^2#

By using algebraic distribution and FOIL, we can reduce this expression to get

#(24x^4-8x^3+4)/(4x-1)^2#

I hope this is useful.

Ezra Adams · Answer

undefined To use the quotient rule to differentiate the function y=(2x^4-3x)/(4x-1), you would follow these steps:

1. Identify the numerator and denominator of the function.
   Numerator: 2x^4 - 3x
   Denominator: 4x - 1

2. Apply the quotient rule, which states that the derivative of a quotient of two functions (u/v) is given by:
   (u'v - uv') / v^2

3. Find the derivatives of the numerator and denominator.
   Derivative of the numerator (u'): 8x^3 - 3
   Derivative of the denominator (v'): 4

4. Substitute these values into the quotient rule formula.
   ( (8x^3 - 3)(4x - 1) - (2x^4 - 3x)(4) ) / (4x - 1)^2

5. Simplify the expression.
   ( (32x^4 - 8x^3 - 12x + 3) - (8x^4 - 12x) ) / (4x - 1)^2
   (24x^4 - 8x^3 + 12x + 3) / (4x - 1)^2

Therefore, the derivative of y=(2x^4-3x)/(4x-1) with respect to x is (24x^4 - 8x^3 + 12x + 3) / (4x - 1)^2.