Question #ab1ff

2 Answers
Dec 19, 2017

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.!

Dec 19, 2017

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:)