We need to differentiate an implicit function. Implicit differentiation is very similar to regular differentiation. We just have to bear in mind that `y` is a function if `x.` Therefore, if we need to differentiate a function of `y` we will proceed as follows:
`d/dx f(y)=(df)/dy cdot dy/dx`
You can see more examples in the link below.
`y=log(xy)`
We differentiate the whole equation (both left and right side).
`dy/dx=1/(xy log10)cdot(y+x cdot dy/dx)`
In the line above we have applied...
We need to differentiate an implicit function. Implicit differentiation is very similar to regular differentiation. We just have to bear in mind that `y` is a function if `x.` Therefore, if we need to differentiate a function of `y` we will proceed as follows:
`d/dx f(y)=(df)/dy cdot dy/dx`
You can see more examples in the link below.
`y=log(xy)`
We differentiate the whole equation (both left and right side).
`dy/dx=1/(xy log10)cdot(y+x cdot dy/dx)`
In the line above we have applied the chain rule because `log(xy)` is a composite function. On top of that we have applied product rule for `xy` to get expression in the brackets.
`dy/dx=1/(x log 10)+1/(y log 10)dy/dx`
`dy/dx-1/(y log 10)dy/dx=1/(x log 10)`
Factor the left side.
`dy/dx(1-1/(y log 10))=1/(x log10)`
`dy/dx cdot (y log 10-1)/(y log 10)=1/(x log 10)`
`dy/dx=y/(x(ylog(10)-1))` <-- Solution
We could also put `log(xy)` instead of `y` but that wouldn't make the solution any simpler.
No comments:
Post a Comment