   Chapter 3.6, Problem 24E

Chapter
Section
Textbook Problem

Find y′ and y″. y = ln x 1 + ln x

To determine

To find: The first and second derivatives of y. That is, y and y.

Explanation

Given:

The function is y=lnx1+lnx.

Derivative Rule: Quotient Rule

If f(x). and g(x) are both differentiable function, then

ddx[fg]=gddx[f]fddx[g](g)2 (1)

Calculation:

Obtain the first derivative of y.

y=ddx(y)=ddx(lnx1+lnx)

Apply the quotient rule as shown in equation (1),

y=(1+lnx)ddx[lnx](lnx)ddx[1+lnx](1+lnx)2=(1+lnx)[1x](lnx)[0+1x](1+lnx)2=1x+lnxxlnxx(1+lnx)2=1x(1+lnx)2

Therefore, the derivative of y=lnx1+lnx is y=1x(1+lnx)2_.

Obtain the second derivative of y.

y=ddx(y)=ddx(1x(1+lnx)2)

Apply the quotient rule as shown in equation (1),

y=x(1+lnx)2ddx(1)ddx[x(1+lnx)2][x(1+lnx)2]2=x(1+lnx)2[xddx

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