Fx-x cosxFind f-Formula- cosA-B- cosA cosB- sinA sinB-

f(x)=x× cosx f'(x)=1× cosx+x×( sinx) =cosx xsinx ∴ f'(π)=cosπ π sinπ = 1 ...

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Parametric Differentiation

fx=x cosxFind...

Question

f(x)=x cosx
Find $f^{'} (π)$
Formula: cos(A+B)= cosA cosB- sinA sinB.

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Solution

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$\frac{s i n A - s i n B}{c o s A + c o s B} + \frac{c o s A - c o s B}{s i n A + s i n B} =$

\frac{sin A - sin B}{cos A + cos B} + \frac{cos A - cos B}{sin A + sin B} = 0

Simplify the expression:
$\frac{s i n A - s i n B}{c o s A + c o s B} + \frac{c o s A - c o s B}{s i n A + s i n B}$

{(\frac{c o s A + c o s B}{s i n A - s i n B})}^{2014} + {(\frac{s i n A + s i n B}{c o s A - c o s B})}^{2014} = . . . . . . . . . . .

{(\frac{cos A + cos B}{sin A - sin B})}^{n} + {(\frac{sin A + sin B}{cos A - cos B})}^{n}

= 2 {cot}^{n} (\frac{A - B}{2})

n

= 0

n

\frac{cos A + cos B}{sin A - sin B} = cot \frac{A - B}{2}

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