Question

Question #ab1ff

Answer 1

`dy/dx={(sinx)^sqrtx(2xcotx+lnsinx)}/(2sqrtx).`

Explanation:

Given that, `y=(sinx)^sqrtx.`

Taking Natural Log. of both sides, and, using `lna^m=mlna,`

`lny=ln{(sinx)^sqrtx}=sqrtx*lnsinx.`

Using the Product Rule to diff. both sides w.r.t. `x,` we get,

`d/dx{lny}=d/dx{sqrtx*lnsinx}, i.e.,`

`d/dx{lny}=sqrtxd/dx{lnsinx}+(lnsinx)d/dxsqrtx....(star).`

Here, by the Chain Rule,

`d/dx{lny}=d/dy{lny}*dy/dx=1/y*dy/dx.`

Similarly, `d/dx{lnsinx}=1/sinx*d/dx{sinx}=cosx/sinx=cotx`.

Utilising these in `(star),` we get,

`1/ydy/dx=sqrtx*cotx+(lnsinx)(1/2*x^(1/2-1)),`

`=sqrtx*cotx+(lnsinx)/(2sqrtx), or,`

`=(2xcotx+lnsinx)/(2sqrtx).`

`rArr dy/dx={y(2xcotx+lnsinx)}/(2sqrtx).`

Since, `y=(sinx)^sqrtx,`

`dy/dx={(sinx)^sqrtx(2xcotx+lnsinx)}/(2sqrtx).`

Enjoy Maths.!

Answer 2

Let's see.

Explanation:

Given `rarr`

`y=(sinx)^(sqrtx)`.......(1).

Taking "log" on both sides,

`color(red)(logy=sqrtxlog(sinx))`

Now differentiating w.r.t `x` by chain rule `rarr`

`:.1/ycdot(dy/dx)=1/(2sqrtx)cdotlog(sinx)+sqrtxcdot(cosx/sinx)`

`:.1/ycdot(dy/dx)=log(sinx)/(2sqrtx) +sqrtxcotx`

`:.dy/dx=(ylogsinx)/(2sqrtx) +ysqrtxcotx`

Now from equation (1) `rarr`

`:.color(red)(dy/dx=((sinx)^(sqrtx)cdotlog(sinx))/(2sqrtx)+((sinx)^(sqrtx)sqrtxcotx))`

`:.color(red)(dy/dx=(sinx)^(sqrtx)(log(sinx)/(2sqrtx)+sqrtxcotx))`

Hope it Helps:)