Question

Differentiate

$\frac{x + cosx}{tanx}$
with respect to
x





wrong =report ​

Answer 1

Answer:

$y = \frac{x + cosx}{tanx}$

By using quotient rule,

$\frac{dy}{dx} = \frac{VU {}^{1} - UV {}^{1} }{V {}^{2} }$

Differnatatiating with respect to X

$\frac{dy}{dx} = \frac{tanx. \frac{d(x + cosx)}{dx} -( x +cosx). \frac{d(tanx)}{dx} }{(tanx) {}^{2} }$

$\frac{dy}{dx } = \frac{tanx.(1 + ( - sinx) - (x + cosx).sec {}^{2} x}{(tanx) {}^{2} }$

$\frac{dy}{dx} = \frac{tanx.(1 - sinx) - sec {}^{2} x(x + cosx) }{ {(tanx)}^{2} }$

Be Brainly!

Answer 2

$\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \:  \: TO \: \: FIND \:  \:  \:  \maltese }}}}} \\ \\ \huge \mathfrak{Derivative \: \: of } \\ \\ \bf \frac{x+cosx}{tanx}$

$\orange{\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \:  \: REQUIRED \: \: INFO \:  \:  \:  \maltese }}}}}$

$\bigstar \: \: \: \mathfrak{ \underline{\huge Quotient \: \: Rule}} \\ \\ \bf \frac{d}{dx} \frac{u}{v} = \frac{v . \frac{du}{dx} - u \frac{dv}{dx} }{v {}^{2} } \\ \\ \bigstar \bf \: \: \: \frac{d}{dx} cosx = - sinx \\ \\ \bigstar \: \: \: \bf \frac{d}{dx} tanx = {sec}^{2} x$

$\green{ \large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \:  \: SOLUTION \:  \:  \:  \maltese }}}}}$

$\bf \frac{d}{dx} ( \frac{x + cosx}{tanx} ) \\ \\ = \sf \frac{tanx. \frac{d}{dx} (x + cosx) - (x + cosx) \frac{d}{dx} tanx}{ {(tanx)}^{2} } \\ ( \bf \because Quotient \: Rule ) \\ \\ = \sf\frac{tanx(1 - sinx) - (x + cosx) {sec}^{2} x}{ {tan}^{2} x}$

$\sf = \frac{tanx \: - \: tanx.sinx \: - \: x {sec}^{2} x \: - \: cosx. {sec}^{2} x}{ {tan}^{2}x } \\ \\ =\sf \frac{tanx \: - \: tanx.sinx \: - \: x {sec}^{2} x \: - \: cosx. \dfrac{1}{ {cos}^{2}x } }{ {tan}^{2}x } \\ \\ = \bf\frac{tanx \: - \: tanx.sinx \: - \: x {sec}^{2} x \: - \: {sec}x}{ {tan}^{2}x }$

$\large\red{ \mathfrak{ \text{W}hich \: \: is \: \text{y}our \: \: Required \: \: \text{ A}nswer }}$