Question

Find dy/dx
where
y=sqrt x^2+x+1/x^2-x+1

​

Answer 1

Answer:

I don't know if you any problem please tell me

Answer 2

Answer:

y = x² + x + 1 / x² - x + 1

taking log on both sides

log y = log [ x² +x +1 / x² - x +1]

log y = log ( x² + x + 1 ) - log ( x² - x + 1)

now differentiate with respect to x

(1/y) dy/dx = 1/x²+x +1 [d/dx (x²+x+1)] - 1/x²-x+1 [d/dx ( x²-x+1)

(1/y) dy/dx = 1/x²+x +1 [ 2x +1 ] - 1/x²-x+1 [ 2x -1]

dy/dx = y [ 1/x²+x +1 (2x +1 ) - 1/x²-x+1 ( 2x -1)]

dy/dx = x² + x + 1 / x² - x + 1 [ 1/x²+x +1 (2x +1 ) - 1/x²-x+1 ( 2x -1)]

hope this will help you

sorry for being late

you can further simplify if it get simplified

(I have solved the correct question or not??)