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

2 Answers
Aug 11, 2018

Answer:

#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

Answer:

#:.(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)#