Question

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Find the derivative of

$(i) \frac{ {x {}^{5} - cosx}^{} }{sinx}$
$(ii) \: \frac{x + cosx}{tanx}$
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Answer 1

Solution:

$\bf 1).\;\dfrac{x^{5}-\cos x}{\sin x}\\ \\ \\ \bf Sol. \;\sf Given: \\ \\ \\ \longrightarrow \sf P = \dfrac{x^{5}-\cos x}{\sin x}\\ \\ \\ \longrightarrow \sf \dfrac{dP}{dx}=\dfrac{\sin x(5x^{4}+\sin x)-(x^{5}-\cos x)\cos x}{\sin^{2} x}\\ \\ \\ \longrightarrow \sf \dfrac{dP}{dx}=\dfrac{5x^{4}\sin x+\sin^{2}x-x^{5}\cos x+\cos^{2} x}{\sin^{2} x}\\ \\ \\ {\boxed {\longrightarrow {\bf \dfrac{dP}{dx}=\dfrac{5x^{4}\sin x-x^{5}\cos x}{\sin^{2}x}}}} \\ \\ \rule{150}{2}$

$\bf 2). \; \dfrac{x+\cos x}{\tan x}\\ \\ \\ \bf Sol. \; \sf Given:\\ \\ \\ \longrightarrow \sf y= \dfrac{x+\cos x}{\tan x}\\ \\ \\ {\underline {\bf Now,\;differentiating\;with\;respect\;to\;'x',\;we\;get,}}\\ \\ \\ \longrightarrow \sf \dfrac{dy}{dx}=\dfrac{\tan x(1-\sin x)-(x+\cos x).\sec^{2} x}{(\tan x)^{2}}\\ \\ \\ \longrightarrow \sf \dfrac{dy}{dx}=\dfrac{\tan x-\dfrac{\sin x}{\cos x}.\sin x-x\sec^{2}x-\cos x.\dfrac{1}{\cos^{2}x}}{(\tan x)^{2}}$

$\longrightarrow \sf \dfrac{dy}{dx}=\dfrac{\tan x-\dfrac{\sin^{2}x}{\cos x}-x\sec^{2}x-\dfrac{1}{\cos x}}{(\tan x)^{2}}\\ \\ \\ \longrightarrow \sf \dfrac{dy}{dx}=\dfrac{\tan x-\dfrac{(\sin^{2}x+1)}{\cos x}-x\sec^{2}x}{(\tan x)^{2}}\\ \\ \\ \longrightarrow \sf \dfrac{dy}{dx}=\dfrac{\dfrac{\sin x}{\cos x}-\dfrac{(\sin^{2}x+1)}{\cos x}\dfrac{x}{\cos^{2}x}}{(\tan x)^{2}}\\ \\ \\ \longrightarrow \sf \dfrac{dy}{dx}=\dfrac{\dfrac{(\sin x-\sin^{2}x-1)\cos x-x}{\cos^{2}x}}{\Bigg(\dfrac{\sin x}{\cos x}\Bigg)^{2}}$

${\boxed {\longrightarrow {\bf \dfrac{dy}{dx}=\dfrac{(\sin x-\sin^{2}x-1)\cos x-x}{\sin^{2} x}}}}$

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