a) Find dy/dx , x^y=y^x b) Use L'Hospitals's Rule to find the limit lim x--> 1 (1-x)/(sinpix- lnx)     

Asked on by hakiki1

2 Answers | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

Notice that the function from denominator is `sin pi*x - ln x`  and the function from numerator is `1 - x,`  hence, if you substitute 1 for x to numerator and denominator yields:

`lim_(x-gt1) (1-x)/(sin pi*x - ln x) = (1-1)/(sin pi - ln 1) = 0/(0-0) = 0/0`

You should use the l'Hospital's theorem, hence you need to differentiate the numerator and denominator with respect to x, as independent functions, such that:

`lim_(x-gt1) ((1-x)')/((sin pi*x - ln x)') = lim_(x-gt1) (-1)/(pi*cos pi*x - 1/x)`

You need to substitute 1 for x such that:

`lim_(x-gt1) (-1)/(pi*cos pi*x - 1/x) = (-1)/(pi*cos pi - 1)`

`lim_(x-gt1) (-1)/(pi*cos pi*x - 1/x) = (-1)/(-pi-1)`

`lim_(x-gt1) (-1)/(pi*cos pi*x - 1/x) = 1/(pi+1) `

Hence, evaluating the limit of the function `(1-x)/(sin pi*x - ln x)`  yields `lim_(x-gt1) (1-x)/(sin pi*x - ln x) = 1/(pi+1).`

thilina-g's profile pic

thilina-g | College Teacher | (Level 1) Educator

Posted on


`x^y = y^x`

using implicit differentiation,

`yx^(y-1) = xy^((x-1))*(dy)/(dx)`

`(dy)/(dx) = (yx^(y-1))/(xy^(x-1))`

`(dy)/(dx) = (x^(y-2))/(y^(x-2))`

We’ve answered 319,824 questions. We can answer yours, too.

Ask a question