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If \( y=x^{\tan x}+(\sin x)^{\cos x} \), Find \( \frac{d y}{d x} \) |
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Answer» y = xtanx + (sinx)cosx Let y1 = xtanx & y2 = (sinx)cosx ∴ \(\frac{dy}{dx}\) = \(\frac{dy_1}{dx}\) + \(\frac{dy_2}{dx}\) .........(1) Now, logy1 = tanx loga & logy2 = cosx log sinx (∵ log ab = blog a) ∴ \(\frac{1}{y_1}\frac{dy_1}{dx}\) = \(\frac{tanx}{x}\) + sec2x logx ⇒ \(\frac{dy_1}{dx}\) = (\(\frac{tanx}{x}\) + sec2x logx) xtanx & \(\frac{1}{y_2}\frac{dy_2}{dx}\) = \(\frac{cosx}{sinx}\times cosx\) - sinx log sinx ⇒ \(\frac{dy_2}{dx}\) = (cosx cotx - sinx log sinx)(sinx)cosx From (1), \(\frac{dy}{dx}\) = (\(\frac{tanx}{x}\) + sec2x logx) xtanx + (cosx cotx - sinx log sinx)(sinx)cosx |
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