Question

If y=x^tanx+(sinx)^cosx find dy/dx?

Answer 1

$\bf y=x^{tanx}+(sinx)^{cosx}$

Apply ln on both sides , we get ,

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

Now , differentiate y with respect to x , we get ,

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

Answer 2

Answer:

=> Refer to the attachment.

Step-by-step explanation:

Plz follow me fast and thnks guys..