How do you find the derivative of $y= arctan(x/a)+ lnsqrt((x-a)/(x+a))$ ?

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Answer 1 · 2017-04-10T20:03:21

$dy/dx = (2ax^2)/(x^4-a^4)$

Using the linearity of the derivative:

$dy/dx = d/dx arctan(x/a) + d/dx ln sqrt((x-a)/(x+a))$

Now:

$d/dx arctan(x/a) = 1/a 1/(1+(x/a)^2) = a/(a^2+x^2)$

and using the properties of logarithms:

$d/dx ln sqrt((x-a)/(x+a)) = d/dx( 1/2(ln (x-a) -ln(x+a)))$

$d/dx ln sqrt((x-a)/(x+a)) = 1/2(1/(x-a) - 1/(x+a))$

$d/dx ln sqrt((x-a)/(x+a)) = 1/2( ((x+a) -(x-a)) /((x-a) (x+a)))$

$d/dx ln sqrt((x-a)/(x+a)) = a/(x^2-a^2)$

Summing the two terms:

$dy/dx = a/(x^2+a^2) + a/(x^2-a^2)$

$dy/dx = (a( x^2-a^2) + a (x^2+a^2) ) / ( (x^2+a^2) (x^2-a^2)) =(2ax^2)/(x^4-a^4)$

How do you find the derivative of y= arctan(x/a)+ lnsqrt((x-a)/(x+a)) y= arctan(x/a)+ lnsqrt((x-a)/(x+a))?