What is dy/dx of a square root using logarithmic differentiation? I thought I knew how to

Camila Glenn

Camila Glenn

Answered question

2022-04-16

What is dydx of a square root using logarithmic differentiation?
I thought I knew how to do logarithmic differentiation until I came across this:
y=x(x+7) find dy/dx by logarithmic differentiation
I was not phased by the problem and did the following work:
ln(y)=ln(x(x+7))
ln(y)=12(ln(x)+ln(x+7))
1ydydx=12(1x+1x+7)
dydx=y2(2x+7x2+7x)
but then found out that that was not actually what the answer was if you take dy/dx without logarithmic differentiation. What am I doing wrong? I have spent the past hour trying to figure it out with no luck.

Answer & Explanation

Coupewopmergorlpt

Coupewopmergorlpt

Beginner2022-04-17Added 13 answers

Looks to me like you've done it correctly. Since y=x2+7x
y2(2x+7x2+7x)=12(2x+7)x2+7xx2+7x=2x+72x2+7x.
This last is of course the derivative (after chain rule) of x2+7x
shanna87mn

shanna87mn

Beginner2022-04-18Added 19 answers

You did everything perfectly. Resubstitute
y=x(x+7)
to get
y=x(x+7)22x+7x(x+7)=2x+72x(x+7)
This is precisely what I get by applying normal differentiation rules.

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