Question

Differentiate the above with clear steps

Answer 1

explanation:

y=tanx^sinx

logy=sinx log tanx

(1/y)(dy/dx) =cosx(log tanx) + sinx(1/tanx) sec^2x

dy/dx=y{cosx(log tanx) +(sinx÷cosx/sinx) 1/cos^2x

dy/dx=y{(cosxlog(tanx) +(1/cosx) }

dy/dx=y{secx+(cosx(logtanx) }

Answer 2

$\large\underline{\sf{Given \:Question - }}$

Differentiate the following function with respect to x

$\rm :\longmapsto\:y = {tanx}^{sinx}$

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

Given function is

$\rm :\longmapsto\:y = {tanx}^{sinx}$

Taking log on both sides, we get

$\rm :\longmapsto\:logy = log \: {tanx}^{sinx}$

We know,

$\boxed{ \bf{ \: log {x}^{y} = y \: logx}}$

So, using this, we get

$\rm :\longmapsto\:logy \: = \: sinx \: log \: tanx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}logy \: = \dfrac{d}{dx}\: sinx \: log \: tanx$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx}logx = \frac{1}{x}}}$

and

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

So, using this result, we get

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{d}{dx}y = sinx\dfrac{d}{dx}logtanx + logtanx\dfrac{d}{dx}sinx$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx}sinx = cosx}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} = sinx\dfrac{1}{tanx}\dfrac{d}{dx}tanx + logtanx \: (cosx)$

We know,

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} = sinx \times \dfrac{cosx}{sinx} \times {sec}^{2}x + logtanx \: (cosx)$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} =cosx \times {sec}^{2}x + logtanx \: (cosx)$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} = {sec}^{}x + logtanx \: (cosx)$

$\rm :\longmapsto\:\dfrac{dy}{dx} =y\bigg[ {sec}^{}x + logtanx \: (cosx)\bigg]$

$\bf :\longmapsto\:\dfrac{dy}{dx} = {tanx}^{sinx} \bigg[ {sec}^{}x + logtanx \: (cosx)\bigg]$

Additional Information :-

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

$\boxed{ \bf{ \: \dfrac{d}{dx}cosecx = - cosecx \: cotx}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}secx = secx \: tanx}}$

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

$\boxed{ \bf{ \: \dfrac{d}{dx}x = 1}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}k = 0}}$