Question 1088245
<pre>
{{{t*ln(x)}}}{{{""=""}}}{{{x*e^t-1 }}}

Differentiating implicitly to find {{{"x'(t)"}}}

{{{t*(1/x)*"x'"+ln(x)}}}{{{""=""}}}{{{x*e^t+e^t*"x'"}}}

Clear the fraction by multiplying through by LCD=x

{{{t*"x'"+x*ln(x)}}}{{{""=""}}}{{{x^2*e^t+x*e^t*"x'"}}}

{{{t*"x'"-x*e^t*"x'"}}}{{{""=""}}}{{{x^2*e^t-x*ln(x)}}}

{{{(t-x*e^t)*"x'"}}}{{{""=""}}}{{{x^2*e^t-x*ln(x)}}}

{{{"x'"=(x^2*e^t-x*ln(x))/(t-x*e^t)}}}

When y is the dependent variable and x is the independent
variable, the point-slope equation of the line through
(x<sub>1</sub>,y<sub>1</sub>) with slope m is {{{y-y[1]}}}{{{""=""}}}{{{m(x-x[1])}}}.

But in this case, x is the dependent variable and t is the independent
variable, so the equation of the line through (t<sub>1</sub>,x<sub>1</sub>)
with slope m is {{{x-x[1]}}}{{{""=""}}}{{{m(t-t[1])}}}.

(t<sub>1</sub>,x<sub>1</sub>) = (0,1)

And the slope m is x' evaluated at this point:

{{{"x'(0,1)"=(1^2*e^0-1*ln(1))/(0-1*e^0)}}}

{{{"x'(0,1)"}}}{{{""=""}}}{{{(1*1-1*0)/(-1*1)}}}

{{{"x'(0,1)"}}}{{{""=""}}}{{{-1}}}
 
So {{{m}}}{{{""=""}}}{{{-1}}}

So substituting m = -1, and (t<sub>1</sub>,x<sub>1</sub>) = (0,1)

{{{x-x[1]}}}{{{""=""}}}{{{m(t-t[1])}}}

{{{x-1}}}{{{""=""}}}{{{-1(t-0)}}}

{{{x-1}}}{{{""=""}}}{{{-t}}}

{{{x}}}{{{""=""}}}{{{-t+1}}}

Edwin</pre>