If u- e logcos 4x and v- e logsin 4x show that dv du - u v

u = e^logcos 4x = cos 4x v = e^logsin 4x = sin 4x dudx = #160; 4sin 4x,dvdx = 4cos 4x dvdu = dvdxdudx = 4cos 4x 4sin 4x = cos ...

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Parametric Differentiation

If u= e log...

Question

If $u = e^{log cos 4 x}$ and $v = e^{log sin 4 x}$ show that $\frac{d v}{d u} = - \frac{u}{v}$

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v = u

\frac{d v}{d u} + {(\frac{d v}{d u})}^{2}

u = y

v = x y

v = u \frac{d v}{d u} + {(\frac{d v}{d u})}^{2}

u = y

v = x y

u

v

x

y = u + v

\frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}

\frac{2}{f} = \frac{1}{v} - \frac{1}{u}

d u = d v

\frac{d f}{f} =

u = f (cot x)

v = g (cosec x)

f^{'} (1) = \sqrt{2}

g^{'} (\sqrt{2}) = 2

{(\frac{d u}{d v})}_{x = \frac{π}{4}} = 1

u = f (x), v = g (x)

f

g

\frac{d u}{d v} = \frac{d u / d x}{d v / d x}

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