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

Name: For the following exercises, use logarithmic differentiation to find dydx y=(x2−1)ln⁡x
Uploaded: 2022-02-23T17:22:57+00:00
Description: For the following exercises, use logarithmic differentiation to find dydx y=(x2−1)ln⁡x

Question

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

Donald Cheek · Accepted Answer

We have given that  y=(x2−1)ln⁡x --- 1  Taking logarithm both side of 1 we get  ln⁡y=ln⁡(x2−1)ln⁡x  ⇒ln⁡y=(ln⁡x)[ln⁡(x2−1)]  Now differentiating both side with respect to x -  ddx(ln⁡y)=ddx[ln⁡x(ln⁡(x2−1))]  ⇒1y⋅dydx=(ln⁡x)ddx[ln⁡x(x2−1)]+ln⁡(x2−1)ddx(ln⁡x)  ⇒1y⋅dydx=(ln⁡x)1x2−1⋅ddx(x2−1)+ln⁡(x2−1)1x  ⇒1y⋅dydx=ln⁡xx2−1⋅(2x)+ln⁡(x2−1)x  ⇒dydx=y[2xln⁡xx2−1+ln⁡(x2−1)x]  ⇒dydx=(x2−1)ln⁡x[2xln⁡xx2−1+ln⁡(x2−1)x]

Corgnatiui · Accepted Answer

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

user_27qwe · Accepted Answer

First answer is right!