Derivative of the function

Question

, y'=?

Answer 1

To make this simpler to discuss let Q = b² - a², N = numerator of the parameter to ln and D = denominator. Note that Q and 1/√Q are constants, and b must be greater then a, otherwise division by 0 or square root of a negative number will result.

I would start by using properties of logarithms to rewrite ln portion as ln N - ln D.

So, y = 1/√Q[ln N - ln D]

Now y' = 1/√Q [ 1/N (dN/dx) - 1/D (dD/dx)]

dN/dx = √Q cosx - asinx

dD/dx = -bsinx

See if you can take if from there.