Question

Use logarithmic differentiation to find dy/dx. y = x^x - 8, x > 0.

Answer 1

You need to know these:

`\sf\\\textsf{1. Chain rule: }\\\\(dy)/(dx)=(dy)/(du)* (du)/(dx)`

`\sf\\\textsf{2. Product rule}\\\\(d(uv))/(dx)=u(dv)/(dx)+v(du)/(dx)`

`\sf\\\textsf{3.} (d(lnx))/(dx)=(1)/(x)`

`\sf\\\textsf{4. }y=x^x\\\textsf{or, }ln(y)=ln(x^x)=xlnx\\\\\textsf{or, }(d(lny))/(dy)=(d(xlnx))/(dx)\\\\\textsf{or, }(d(lny))/(dy)*(dy)/(dx)=x(d(lnx))/(dx)+lnx(dx)/(dx)`

`\sf\\\textsf{or, }(1)/(y)* (dy)/(dx)=x(1)/(x)+lnx=1+lnx\\\\\textsf{or, }(dy)/(dx)=y(1+lnx)\\\\\textsf{Substitute }y=x^x\\\textsf{}\ \ \ \ (d(x^x))/(dx)=x^x(1+lnx)`

Answer:

`\sf\\y=x^x-8\\`

`\textsf{Applying natural log on both sides,}`

`\sf\\ln(y)=ln(x^x-8)`

`\textsf{Differentiating both sides with respect to x,}`

`\sf\\(d(lny))/(dx)=(d[ln(x^x-8)])/(dx)`

`\sf\\\textsf{or, }(d(lny))/(dy)*(dy)/(dx)=(d[ln(x^x-8)])/(d(x^x-8))*(d(x^x-8))/(dx^x)*(dx^x)/(dx)`

`\sf\\\textsf{or, }(1)/(y)* (dy)/(dx)=(1)/(x^x-8)*1* x^x(1+lnx)`

`\sf\\\textsf{or, }(1)/(y)* (dy)/(dx)=(x^x(1+lnx))/(x^x-8)\\`

`\sf\\\textsf{or, }(dy)/(dx)=y*(x^x(1+lnx))/(x^x-8)`

`\sf\\\textsf{or, }(dy)/(dx)=((x^x-8)[x^x(1+lnx)])/(x^x-8)\\`

`\sf\\\therefore\ (dy)/(dx)=x^x(1+lnx)`