Help! Would this be the correct answer to the following logarithmic differentiation problem?

Question

Use logarithmic differentiation to find the derivative of the function y= (2x+5)³(5x²-2x)⁴

———————

√2x+5

the answer that I got was

dy/dx= ((2x+5)³(5x²-2x)⁴/√2x+5) (6/2x+5 + 40+240x³/3x - 1/2x+5)

would that work/be correct? thanks!

Answer 1

y= (2x+5)³(5x²-2x)⁴

———————

√2x+5

means y= (2x+5)^5/2(5x²-2x)⁴ because there is the division by √2x+5

then ln(y)=(5/2)ln(2x+5)+4ln(5x²-2x)

[(dy/dy)/y]=[(5/2)(2)/(2x+5)]+[4(10x-2)/(5x²-2x)]

dy/dx=([(5/2)(2)/(2x+5)]+[4(10x-2)/(5x²-2x)])([(2x+5)³(5x²-2x)⁴ ])

the result should be two terms, and the denominators divide into numerators.

is that how you proceeded. it is checkable by simply using product rule.

dy/dx=(5/2)(2)(2x+5)^3/2(5x²-2x)⁴+(2x+5)^5/2(4)(10x-2)(5x²-2x)³

Answer 2

Hi Anne.

y = ((2x+5)³(5x²-2x)⁴)/√(2x+5)

Take the natural log of each side:

ln(y) = ln(((2x+5)³(5x²-2x)⁴)/√(2x+5))

Perform log rule steps on the right side:

ln(y) = ln((2x+5)³) + ln((5x²-2x)⁴) - ln(√(2x+5))

ln(y) = 3*ln(2x+5) + 4*ln(5x²-2x) - 0.5*ln(2x+5)

Implicitly differentiate each side. The derivative of natural log is "one over whatever is inside the log, times the derivative of whatever's inside the log".

1/y * dy/dx = 3*1/(2x+5)*2 + 4*1/(5x²-2x)*(10x-2) - 0.5*1/(2x+5)*2

1/y * dy/dx = 6/(2x+5) + (40x-8)/(5x²-2x) - 1/(2x+5)

dy/dx = y * (6/(2x+5) + (40x-8)/(5x²-2x) - 1/(2x+5))

dy/dx = ((2x+5)³(5x²-2x)⁴)/√(2x+5) * (6/(2x+5) + (40x-8)/(5x²-2x) - 1/(2x+5))

Looks like the only difference was that middle term of the implicit differentiation, otherwise you got everything else right!