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

Posted in Math, asked by wisdom bruce, 6 years ago. 2823 hits.

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Please refer to the attachment   given below

Sankar Rao - 6 years ago
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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)

Russell Ludlow - 6 years ago
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