Q-y-xcosx-x-xsinx-1-x-then-find-dy-dx-

Question Number 112720 by deepak@7237 last updated on 09/Sep/20

Q . y = {(x \cos x)}^{x} + {(x \sin x)}^{\frac{1}{x}} then find \frac{d y}{d x}

Answered by deepak@7237 last updated on 09/Sep/20

Answered by 1549442205PVT last updated on 09/Sep/20

Put f (x) = {(xcosx)}^{x}, g (x) = {(xsinx)}^{\frac{1}{x}}

\ln [f (x)] = xln (xcosx), \ln [g (x)] = \frac{1}{x} \ln (xsinx)

Derivative two sides of each equality

by x we get :

\frac{f^{'} (x)}{f (x)} = \ln (xcosx) + x . [\frac{1}{xcosx} \times (cosx - xsinx)]

= \ln (xcosx) + \frac{cosx - xsinx}{cosx}

\Rightarrow f^{'} (x) = {(xcosx)}^{x} \times [\ln (xcosx) + \frac{cosx - xsinx}{cosx}]

\frac{g^{'} (x)}{g (x)} = (- \frac{1}{x^{2}}) \ln (xsinx) + \frac{1}{x} [\frac{1}{xsinx} \times (sinx + xcosx)]

= - \frac{1}{x^{2}} \ln (xsinx) + \frac{sinx + xcosx}{x^{2} sinx}

\Rightarrow g^{'} (x) = {(xsinx)}^{\frac{1}{x}} \times [\frac{1}{x^{2}} \ln (xsinx) + \frac{sinx + xcosx}{x^{2} sinx}]

\frac{dy}{dx} = f^{'} (x) + g^{'} (x)

= {(xcosx)}^{x} \times [\ln (xcosx) + \frac{cosx - xsinx}{cosx}]

+ {(xsinx)}^{\frac{1}{x}} \times [\frac{1}{x^{2}} \ln (xsinx) + \frac{sinx + xcosx}{x^{2} sinx}]