differentiae with inspect to x lin{cosx+sinx/cosx-sinx}

Please refer to the attachment   given below

By lin do you mean ln, the natural log? 

To solve this question we need to use the chain rule (for a function inside a function, in this case {cosx+sinx/cosx-sinx} inside ln) 

and the quotient rule (for a function over another function, in this case cosx+sinx over cosx-sinx), plus know the derivative of ln(x), cos x and sin x.

Chain rule: d/dx f(g(x)) = f'(g(x))*g'(x) , or 'differentiate the "inside function" and "outside function" separately then multiply'.

Quotient rule: d/dx f(x)/g(x) = {f'(x)g(x) - f(x)g'(x)} /(g(x))², or 'differentiated top' times bottom minus 'differentiated bottom' times top, all over 'bottom' squared.

d/dx sin(x) = cos(x) , d/dx cos(x) = -sin(x) , d/dx ln(x) = 1/x.

Putting this all together,  d/dx ln{cosx+sinx/cosx-sinx} = 1/{cosx+sinx/cosx-sinx}*(d/dx {cosx+sinx/cosx-sinx})

d/dx {cosx+sinx/cosx-sinx} = {(cosx-sinx)*(cosx-sinx) - (cosx+sinx)*(-sinx-cosx)}/(cosx-sinx)²

now, expanding the brackets and using cos²x+sin²x = 1, we can simplify this to 2/(cosx-sinx)²

Also, from above, 1/{cosx+sinx/cosx-sinx}= (cosx-sinx)/(cosx+sinx), so now the problem is

{(cosx-sinx)/(cosx+sinx)}*{2/(cosx-sinx)²} = 2/(cosx+sinx)(cosx-sinx) = 2/(cos²x-sin²x)

And we are finally done! So d/dx ln{cosx+sinx/cosx-sinx} = 2/(cos²x-sin²x)

Hope this helps! Let me know if any part of this is confusing (I wouldn't be surprised if it was! Try to follow the steps for yourself on paper)