`ln(xy) + 5x = 30` Use implicit differentiation to find dy/dx

Find `"dy"/"dx" ` if `ln(xy)+5x=30 ` :

Use a property of logarithms to rewrite the first term:

`lnx+lny+5x=30 `

Now differentiate term by term with respect to x:

`1/x+1/y*(dy)/(dx)+5=0 `

`1/y*(dy)/(dx)=-(1/x+5) `

`(dy)/(dx)=-y(1/x+5) `

or `(dy)/(dx)=-(y(1+5x))/x `

Find `"dy"/"dx" ` if `ln(xy)+5x=30 ` :

Use a property of logarithms to rewrite the first term:

`lnx+lny+5x=30 `

Now differentiate term by term with respect to x:

`1/x+1/y*(dy)/(dx)+5=0 `

`1/y*(dy)/(dx)=-(1/x+5) `

`(dy)/(dx)=-y(1/x+5) `

or `(dy)/(dx)=-(y(1+5x))/x `