Question

Find d/dx [e^x logx/sin 2x]

Answer 1

Answer:

hope this helps,.........................

Answer 2

$\frac{d(\frac{e^xlogx}{sin2x})}{dx}=\frac{e^xlogx(sin2x-2cos2x)+\frac{e^xsin2x}{x}}{sin^22x}$

Step-by-step explanation:

$\frac{d(\frac{e^xlog x}{sin 2x}}{dx}$

We know that

$\frac{d(u\cdot v)}{dx}=u'v+v'u$...(1)

$\frac{d(\frac{u}{v})}{dx}=\frac{u'v-v'u}{v^2}$...(2)

u$=e^x,v=logx$

By using the formula (1)

$\frac{d(e^xlogx)}{dx}=e^xlogx+e^x\times\frac{1}{x}=e^xlogx+\frac{e^x}{x}$

Formula used $\frac{d(e^x)}{dx}=e^x,\frac{d(logx)}{dx}=\frac{1}{x}$

$u=e^xlogx,v=sin2x$

By formula (2)

$\frac{d(\frac{e^xlogx}{sin2x})}{dx}=\frac{(e^xlogx+\frac{e^x}{x})sin2x-2cos2x\times e^xlogx}{(sin2x)^2}$

$\frac{d(\frac{e^xlogx}{sin2x})}{dx}=\frac{e^xsin2xlogx+\frac{e^xsin2x}{x}-2cos2xe^xlogx}{sin^22x}$

Formula used$\frac{d(sinx)}{dx}=cosx$

$\frac{d(\frac{e^xlogx}{sin2x})}{dx}=\frac{e^xlogx(sin2x-2cos2x)+\frac{e^xsin2x}{x}}{sin^22x}$

#Learns more:

Y=sin^2x×logx ...differentiate with respect to d÷dx​

https://brainly.in/question/9425316:answered by Sk2