Use logarithmic differentiation to find y’ : (question in description)

Question

y=(⁵√x4)cosx ÷ 3(2x+6) simplify answer 

Answer 1

 

 ln y = (4/5) ln x + ln cosx - ln6 - ln (x + 3)

 

 dy/y =  4/(5x) - sinx/cosx - 1/(x+3)

 

 

Multiplies both side by the original function y:

 

 dy  = [4/(5x) - sinx/cosx - 1/(x+3)] [ x^(4/5) cos x / (6(x+3))

 

           = 4 cosx/(30 * x^(1/5) (x+3) - x^(4/5)sinx / (6(x+3)) -  x^(4/5)cosx  / 6(x+3)^2

 

          = (2/15)cosx/ (x^(1/5)(x+3) - (x^(4/5)sinx / (6(x+3)) - (x^(4/5)cosx)/(6(x+3)^2)

 

        =    (4/5)cosx(x+3) - x(sinx)(x+3) - x^(4/5)(cosx) / (6 (x^(1/5))(x+3)^2)

 

       =  (x+3) [(4/5) cosx - x sin x) ] - x^(4/5) cos x /   (6 x^(1/5)(x+3)^2)

 

  which is the same as if the quotient rule were applied  

Answer 2

Hi Lexie,

 

The idea behind logarithmic differentiation is to simplify finding y' when complicated product/quotient rule is required.

Take the natural log of both sides, use properties of logarithms to expand the ln f(x) into separate terms, then take the derivative of both sides. Finally multiply both sides by y (or f(x)) to get y'.

 

In this problem here is what we get after we take the log of both sides and expand the right side:

 

ln y = 4/5 (ln x) + ln (cosx) - ln (6x+18)

 

Now take the derivative of both sides with respect to x.

 

y'/y  = 4/5(1/x) - sinx/cosx - 1/(x+3)

 

Finally multiply both sides by y (or x^4/5cosx/(6x+18)) to get:

 

y' = x^4/5 cosx/(6x+18) [ 4/5x - tanx - 1/(x+3)]

 

In this case this is a tad easier than applying the quotient/product rule to the original function. In some cases logarithmic differentiation wins hands down.