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

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

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Answer 1 · 2018-08-11T17:59:05

$\frac{d y}{d x} = e^{\sqrt{\frac{1}{x} - x^{2}}} . ⎛ ⎜ ⎜ ⎝ \frac{1}{2 \sqrt{\frac{1}{x} - x^{2}}} (- \frac{1}{x^{2}} - 2 x) ⎞ ⎟ ⎟ ⎠$ $dy/dx = e^(sqrt(1/x - x^2)) . (1/(2sqrt(1/x - x^2))(-1/x^2-2x))$

Explanation:

Given,
$f (x) = e^{\sqrt{\frac{1}{x^{2}} - x}}$ $f(x)=e^sqrt(1/x^2 -x)$
Let
$y = f (x)$ $y=f(x)$
Then,
$y = e^{\sqrt{\frac{1}{x^{2}} - x}}$ $y=e^sqrt(1/x^2 -x)$
Let,
$u = \sqrt{\frac{1}{x} - x^{2}}$ $u=sqrt(1/x - x^2)$
Then,
$y = e^{u}$ $y=e^u$
Differentiating both sides, wrt x

$\frac{d y}{d x} = e^{u} . \frac{d u}{d x}$ $dy/dx = e^u . (du)/(dx)$

Now,
$u = \sqrt{\frac{1}{x} - x^{2}}$ $u=sqrt(1/x - x^2)$

Let,
$v = \frac{1}{x} - x^{2}$ $v=1/x - x^2$
Then,
$u = \sqrt{v}$ $u=sqrtv$
Differentiating both sides, wrt x

$\frac{d u}{d x} = \frac{1}{2 \sqrt{v}} \frac{d v}{d x}$ $(du)/(dx) = 1/(2sqrtv)(dv)/(dx)$

Thus,

$\frac{d y}{d x} = e^{u} \frac{1}{2 \sqrt{v}} \frac{d v}{d x}$ $dy/dx = e^u 1/(2sqrtv)(dv)/(dx)$
Again,

$v = \frac{1}{x} - x^{2}$ $v =1/x - x^2$
Let
$r = \frac{1}{x}$ $r = 1/x$
Differentiating wrt x
$\frac{d v}{d x} = - \frac{1}{x^{2}}$ $(dv)/(dx)=-1/x^2$
$s = x^{2}$ $s = x^2$
Differentiating wrt x
$\frac{d s}{d x} = 2 x$ $(ds)/(dx)=2x$
now,
$v = r - s$ $v=r-s$

Differentiating wrt x
$\frac{d v}{d x} = \frac{d r}{d x} - \frac{d s}{d x}$ $(dv)/(dx)=(dr)/(dx)-(ds)/(dx)$
Thus,
$\frac{d v}{d x} = - \frac{1}{x^{2}} - 2 x$ $(dv)/(dx)=-1/x^2-2x$
Substituting for $v & \frac{d v}{d x}$ $v & (dv)/(dx)$ in $\frac{d u}{d x}$ $(du)/(dx)$

$\frac{d u}{d x} = \frac{1}{2 \sqrt{v}} \frac{d v}{d x}$ $(du)/(dx) = 1/(2sqrtv)(dv)/(dx)$
$\frac{d u}{d x} = \frac{1}{2 \sqrt{\frac{1}{x} - x^{2}}} (- \frac{1}{x^{2}} - 2 x)$ $(du)/(dx) = 1/(2sqrt(1/x - x^2))(-1/x^2-2x)$

Substituting for $u & \frac{d u}{d x}$ $u & (du)/(dx)$ in $\frac{d y}{d x}$ $dy/dx$

$\frac{d y}{d x} = e^{u} . \frac{d u}{d x}$ $dy/dx = e^u . (du)/(dx)$

$\frac{d y}{d x} = e^{\sqrt{\frac{1}{x} - x^{2}}} . ⎛ ⎜ ⎜ ⎝ \frac{1}{2 \sqrt{\frac{1}{x} - x^{2}}} (- \frac{1}{x^{2}} - 2 x) ⎞ ⎟ ⎟ ⎠$ $dy/dx = e^(sqrt(1/x - x^2)) . (1/(2sqrt(1/x - x^2))(-1/x^2-2x))$

Answer 2 · 2018-08-11T18:14:11

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

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. $v=1/x^2-x$

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

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

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