For the following exercises, use logarithmic differentiation to find dydx y=(x2−1)ln⁡x

William Curry

William Curry

Answered

2021-12-25

For the following exercises, use logarithmic differentiation to find dydx
y=(x21)lnx

Answer & Explanation

Donald Cheek

Donald Cheek

Expert

2021-12-26Added 41 answers

We have given that
y=(x21)lnx --- 1
Taking logarithm both side of 1 we get
lny=ln(x21)lnx
lny=(lnx)[ln(x21)]
Now differentiating both side with respect to x -
ddx(lny)=ddx[lnx(ln(x21))]
1ydydx=(lnx)ddx[lnx(x21)]+ln(x21)ddx(lnx)
1ydydx=(lnx)1x21ddx(x21)+ln(x21)1x
1ydydx=lnxx21(2x)+ln(x21)x
dydx=y[2xlnxx21+ln(x21)x]
dydx=(x21)lnx[2xlnxx21+ln(x21)x]
Corgnatiui

Corgnatiui

Expert

2021-12-27Added 35 answers

Possible derivation:
ddx((1+x2)log(x))
Express (x21)log(x) as a power of e:(x21)log(x)=elog((x21)log(x))=elog(x)log(x21):
=ddx(elog(x)log(1+x2))
Using the chain rule, ddx(elog(x)log(x21))=deudududx, where u=log(x)log(x21) and ddu(eu)=eu:
=(ddx(log(x)log(1+x2)))elog(x)log(1+x2)
Express elog(x)log(x21) as a power of x:elog(x)log(x21)=elog(xlog(x21))=xlog(x21):
=(xlog(1+x2))ddx(log(x)log(1+x2))
Use the product rule, ddx(uv)=vdudx+ud
user_27qwe

user_27qwe

Expert

2021-12-30Added 230 answers

First answer is right!

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?