how does mirabell introduce lady wishfort in The way of the world by william congreve?
Answers
Answer:
Lady Wishfort offers Mirabell her consent to the marriage if he can save her fortune and honour. Mirabell calls on Waitwell who brings a contract from the time before the marriage of the Fainalls in which Mrs. Fainall gives all her property to Mirabell.
Explanation:
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Answer:
\begin{gathered} \red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: \: TO \: \: FIND \: \: \: \maltese }}}}} \\ \\ \huge \mathfrak{Derivative \: \: of } \\ \\ \bf \frac{x+cosx}{tanx} \end{gathered}✠ TOFIND ✠✠ TOFIND ✠Derivativeoftanxx+cosx
\color{Cyan}{\qquad \qquad \large \underline{ \pmb{{ \mathbb{ \maltese \: REQUIRED \: \: INFO \: \maltese }}}}}✠ REQUIREDINFO ✠✠ REQUIREDINFO ✠
\begin{gathered} \bigstar \: \: \: \mathfrak{ \underline{\huge Quotient \: \: Rule}} \\ \\ \bf \frac{d}{dx} \frac{u}{v} = \frac{v . \frac{du}{dx} - u \frac{dv}{dx} }{v {}^{2} } \\ \\ \bigstar \bf \: \: \: \frac{d}{dx} cosx = - sinx \\ \\ \bigstar \: \: \: \bf \frac{d}{dx} tanx = {sec}^{2} x\end{gathered}★QuotientRuledxdvu=v2v.dxdu−udxdv★dxdcosx=−sinx★dxdtanx=sec2x
\color{magenta}{ \large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: \: SOLUTION \: \: \: \maltese }}}}}✠ SOLUTION ✠✠ SOLUTION ✠
\begin{gathered} \bf \frac{d}{dx} ( \frac{x + cosx}{tanx} ) \\ \\ = \sf \frac{tanx. \frac{d}{dx} (x + cosx) - (x + cosx) \frac{d}{dx} tanx}{ {(tanx)}^{2} } \\ ( \bf \because Quotient \: Rule ) \\ \\ = \sf\frac{tanx(1 - sinx) - (x + cosx) {sec}^{2} x}{ {tan}^{2} x} \end{gathered}dxd(tanxx+cosx)=(tanx)2tanx.dxd(x+cosx)−(x+cosx)dxdtanx(∵QuotientRule)=tan2xtanx(1−sinx)−(x+cosx)sec2x
\begin{gathered} \sf = \frac{tanx \: - \: tanx.sinx \: - \: x {sec}^{2} x \: - \: cosx. {sec}^{2} x}{ {tan}^{2}x } \\ \\ =\sf \frac{tanx \: - \: tanx.sinx \: - \: x {sec}^{2} x \: - \: cosx. \dfrac{1}{ {cos}^{2}x } }{ {tan}^{2}x } \\ \\ = \bf\frac{tanx \: - \: tanx.sinx \: - \: x {sec}^{2} x \: - \: {sec}x}{ {tan}^{2}x } \end{gathered}=tan2xtanx−tanx.sinx−xsec2x−cosx.sec2x=tan2xtanx−tanx.sinx−xsec2x−cosx.cos2x1=tan2xtanx−tanx.sinx−xsec2x−secx
\large\red{ \mathfrak{ \text{W}hich \: \: is \: \text{T}he \: \: Required \: \: \text{ A}nswer }}WhichisTheRequired Answer