Question

How do you differentiate `f(x)=e^sqrt(1/x^2-x)` using the chain rule.?

Answer 1

`dy/dx = e^(sqrt(1/x - x^2)) . (1/(2sqrt(1/x - x^2))(-1/x^2-2x))`

Explanation:

Given,
`f(x)=e^sqrt(1/x^2 -x)`
Let
`y=f(x)`
Then,
`y=e^sqrt(1/x^2 -x)`
Let,
`u=sqrt(1/x - x^2)`
Then,
`y=e^u`
Differentiating both sides, wrt x

`dy/dx = e^u . (du)/(dx)`

Now,
`u=sqrt(1/x - x^2)`

Let,
`v=1/x - x^2`
Then,
`u=sqrtv`
Differentiating both sides, wrt x

`(du)/(dx) = 1/(2sqrtv)(dv)/(dx)`

Thus,

`dy/dx = e^u 1/(2sqrtv)(dv)/(dx)`
Again,

`v =1/x - x^2`
Let
`r = 1/x`
Differentiating wrt x
`(dv)/(dx)=-1/x^2`
`s = x^2`
Differentiating wrt x
`(ds)/(dx)=2x`
now,
`v=r-s`

Differentiating wrt x
`(dv)/(dx)=(dr)/(dx)-(ds)/(dx)`
Thus,
`(dv)/(dx)=-1/x^2-2x`
Substituting for `v & (dv)/(dx)` in `(du)/(dx)`

`(du)/(dx) = 1/(2sqrtv)(dv)/(dx)`
`(du)/(dx) = 1/(2sqrt(1/x - x^2))(-1/x^2-2x)`

Substituting for `u & (du)/(dx)` in `dy/dx`

`dy/dx = e^u . (du)/(dx)`

`dy/dx = e^(sqrt(1/x - x^2)) . (1/(2sqrt(1/x - x^2))(-1/x^2-2x))`

Answer 2

`:.(dy)/(dx)=-1/2(2/x^3+1)e^sqrt(1/x^2-x)/sqrt(1/x^2-x)`

Explanation:

Here ,

`f(x)=y=e^sqrt(1/x^2-x)`

Let ,

`y=e^u` , `u=sqrtv and v=1/x^2-x=x^-2-x`

`(dy)/(dx)=e^u` , `(du)/(dv)=1/(2sqrtv) and (dv)/(dx)=-2x^-3-1=(-2)/x^3-1`

Using Chain Rule:

`color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dv)(dv)/(dx)`

`(dy)/(dx)=e^u 1/(2sqrtv)(-2/x^3-1)`

Subst. back `u=sqrtv`

`(dy)/(dx)=e^sqrtv 1/(2sqrtv)(-2/x^3-1)`

Now ,subst. `v=1/x^2-x`

`:.(dy)/(dx)=e^sqrt(1/x^2-x)1/(2sqrt(1/x^2-x))(-2/x^3-1)`

`:.(dy)/(dx)=-1/2(2/x^3+1)e^sqrt(1/x^2-x)/sqrt(1/x^2-x)`