Question

y=(cotx)^sinx+(tanx)^cosx.find its dy/dx.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: y = {(cotx)}^{sinx} + {(tanx)}^{cosx}$

Let assume that

$\rm \: y = u + v$

where,

$\rm \: u = {(cotx)}^{sinx} - - - (1)$

and

$\rm \: v = {(tanx)}^{cosx} - - - (2)$

So,

$\rm \: \dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dv}{dx} - - - - (3)$

Now, Consider

$\rm \: u = {(cotx)}^{sinx}$

On taking log on both sides, we get

$\rm \:log u = log{(cotx)}^{sinx}$

can be rewritten as

$\rm \:log u = sinx \: logcotx$

On differentiating both sides w. r. t. x, we get

$\rm \:\dfrac{d}{dx}log u = \dfrac{d}{dx}[sinx \: logcotx]$

$\rm \: \dfrac{1}{u}\dfrac{du}{dx} = sinx\dfrac{d}{dx}logcotx + logcotx\dfrac{d}{dx}sinx$

$\rm \: \dfrac{1}{u}\dfrac{du}{dx} = sinx \times \dfrac{1}{cotx}( - {cosec}^{2} x) + logcotx \times cosx$

$\rm \: \dfrac{du}{dx} =u[ - secx \: + \: cosx \: logcotx]$

$\rm\implies \:\: \dfrac{du}{dx} ={(cotx)}^{sinx}[ - secx \: + \: cosx \: logcosx] - - (4)$

Now, Consider

$\rm \: v = {(tanx)}^{cosx}$

On taking log on both sides, we get

$\rm \:log v =log {(tanx)}^{cosx}$

$\rm \:log v =cosx \: log {(tanx)}$

On differentiating both sides w. r. t. x, we get

$\rm \:\dfrac{d}{dx}log v =\dfrac{d}{dx}[cosx \: log {(tanx)}]$

$\rm \: \dfrac{1}{v}\dfrac{dv}{dx} = cosx\dfrac{d}{dx}log(tanx) + log(tanx)\dfrac{d}{dx}cosx$

$\rm \: \dfrac{1}{v}\dfrac{dv}{dx} = cosx \times \dfrac{1}{tanx} \times {sec}^{2}x + log(tanx)( - sinx)$

$\rm \: \dfrac{dv}{dx} = v[cosecx - sinx \: log(tanx)]$

$\rm\implies \: \dfrac{dv}{dx} = {(tanx)}^{cosx}[cosecx - sinx \: log(tanx)] - - - (5)$

So, on substituting the values from equation (4) and (5) in equation (3), we get

$\rm \: \dfrac{dy}{dx} = {(cotx)}^{sinx}[ - secx \: + \: cosx \: logcosx] + {(tanx)}^{cosx}[cosecx - sinx \: log(tanx)]$

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FORMULAE USED

$\rm \: \: \dfrac{d}{dx}tanx = {sec}^{2}x$

$\rm \: \: \dfrac{d}{dx}cotx = - {cosec}^{2}x$

$\rm \: \: \dfrac{d}{dx}sinx = cosx$

$\rm \: \: \dfrac{d}{dx}cosx = - sinx$

$\rm \: \: \dfrac{d}{dx}logx = \dfrac{1}{x}$

$\rm \: log {x}^{y} \: = \: y \: logx$

$\rm \: \dfrac{d}{dx}uv = v\dfrac{d}{dx}u \: + \: u\dfrac{d}{dx}v$

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$