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

2 Answers
Aug 11, 2018

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))

Aug 11, 2018

:.(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)