Question

if y=x^tan x +(sin x)^cos x, find dy/dx​

Answer 1

Answer:

$\tt \dfrac{dy}{dx} =x^{tan\:x}\times\bigg(log\:x\:sec^{2}\:x+\dfrac{tan\:x}{x}\bigg)+ sin\:x^{cos\:x}\times\bigg(-sin\:x\:log\:sin\:x+\dfrac{cos^{2} \:x}{sin\:x }\bigg)$

Step-by-step explanation:

Given:

$\tt y=x^{tan\:x} + sin\:x^{cos\:x}$

To Find:

$\tt \dfrac{dy}{dx}$

Solution:

$\tt \dfrac{dy}{dx} =\dfrac{d}{dx} (x^{tan\:x}+sin\:x^{cos\:x})$

$\tt \implies \dfrac{d}{dx} x^{tan\:x}+\dfrac{d}{dx} sin\:x^{cos\:x}$

Now let,

$\tt u = x^{tan\:x}$

$\tt v= sin\:x^{cos\:x}$

Hence,

$\tt \dfrac{dy}{dx} =\dfrac{du}{dx} +\dfrac{dv}{dx}---(1)$

$\tt u =x^{tan\:x}$

Taking log on both sides,

$\tt log\:u=log\: x^{tanx}$

We know,

$\tt log\:a^m=m\:log\:a$

Hence,

$\tt log\:u=tan\:x\:log\:x$

Differentiate on both sides with respect to x,

$\implies \tt \dfrac{1}{u}\: \dfrac{du}{dx} =log\:x\times sec^{2}\:x+\dfrac{1}{x}\:tan\:x$

Since,

$\tt \dfrac{d}{dx} (log\:x)=\dfrac{1}{x}$

$\tt \dfrac{d}{dx}(uv)=u'v+v'u$

$\tt \dfrac{d}{dx} (tan\:x)=sec^{2}\:x$

Hence,

$\tt \dfrac{du}{dx} =u\times\bigg(log\:x\:sec^{2}\:x+ \dfrac{tan\:x}{x}\bigg)$

Substituting the value of u,

$\tt \dfrac{du}{dx} =x^{tan\:x}\times\bigg(log\:x\:sec^{2}\:x+ \dfrac{tan\:x}{x}\bigg)---(2)$

Now,

$\tt v=sin\:x^{cos\:x}$

Taking log on both sides,

$\tt log\:v=cos\:x\:log\:sin\:x$

Differentiating with respect to x,

$\tt \dfrac{1}{v}\: \dfrac{dv}{dx} =log\:sin\:x\times -sin\:x+\dfrac{1}{sin\:x} \times cos\:x\times cos\:x$

$\tt \dfrac{dv}{dx} =v\times \bigg(-sin\:x\:log\:sin\:x+\dfrac{cos^2\:x}{sin\:x} \bigg)$

Substitute value of v,

$\tt \dfrac{dv}{dx} =sin\:x^{cos\:x}\times \bigg(-sin\:x\:log\:sin\:x+\dfrac{cos^2\:x}{sin\:x} \bigg)----(3)$

Now substitute 2 and 3 in 1,

$\tt \dfrac{dy}{dx} =x^{tan\:x}\times\bigg(log\:x\:sec^{2}\:x+\dfrac{tan\:x}{x}\bigg)+ sin\:x^{cos\:x}\times\bigg(-sin\:x\:log\:sin\:x+\dfrac{cos^{2} \:x}{sin\:x }\bigg)$

Answer 2

Given:-

• $\bf y=x^{tanx}+(sinx)^{cosx}$

To Find:-

• $\bf\dfrac{dy}{dx}$

Solution:-

Apply ln on both sides ,

$\implies \sf ln(y)=ln(x^{tanx})+ln(sinx^{cosx})\\\\\implies \sf ln(y)=tanx.lnx+cosx.ln(sinx)\;[\;Since\ ln(a^b)=b.lna]$

Now , differentiate y with respect to x ,

$\implies \sf \dfrac{d}{dx}(ln(y))=\dfrac{d}{dx}(tanx.lnx+cosx.ln(sinx))\\\\\implies \sf \dfrac{1}{y}.\dfrac{dy}{dx}=\dfrac{d}{dx}(tanx.lnx)+\dfrac{d}{dx}(cosx.ln(sinx))\;[\;Since\;,\dfrac{d}{dx}(lnx)=\dfrac{1}{x\;}]\\\\\implies \sf \dfrac{1}{y}.\dfrac{dy}{dx}=lnx.\dfrac{d}{dx}(tanx)+tanx\dfrac{d}{dx}(lnx)+ln(sinx)\dfrac{d}{dx}(cosx)+cosx\dfrac{d}{dx}(ln(sinx))$

$\sf Since\;,\dfrac{d}{dx}(uv)=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\\\\\implies \sf \dfrac{1}{y}.\dfrac{dy}{dx}=lnx.sec^2x+\dfrac{tanx}{x}+ln(sinx)(-sinx)+cosx.\dfrac{1}{sinx}.cosx\\\\\implies \bf \dfrac{dy}{dx}=y(lnx.sec^2x+\dfrac{tanx}{x}-sinx.ln(sinx)+cotx.cosx)\\\\\implies \sf \dfrac{dy}{dx}=(x^{tanx}+(sinx)^{cosx})(lnx.sec^2x+\dfrac{tanx}{x}-sinx.ln(sinx)+cotx.cosx)$

Answer:-

$\bf \dfrac{dy}{dx}=(x^{tanx}+(sinx)^{cosx})(lnx.sec^2x+\dfrac{tanx}{x}-sinx.ln(sinx)+cotx.cosx)$

★ Hope it helps u ★