Question

y= root sin root x ,then dy/dx=.....

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Answer 1

Answer:

y=sin(x−−√)

Let x−−√=u

When taking the derivative of a function whose input is another function, you must use the chain rule:

dydx=dydu∗dudx

In this case, dydu=d(sin(u))du=cos(u), and dudx=x√dx=12x√. This makes the answer dydx=cosu∗12x√=cos(x√)2x√

How do you differentiate y=root (sin (root x))?

What is the derivative of sin(x−−√)−−−−−−−√?

How do I solve (x−2ey)dy+(y+xsinx)dx=0 ?

If y=sin x^0 then find dy/dx.

If y=x, what is dy/dx?

y= sin√x

For finding dy/dx just follow the following steps.

Let √x =t

Now y=sint

We have to find dy/dx but now we have a function of t. So we can rewrite dy/dx as..

dy/dx =(dy/dt)(dt/dx)

Now dy/dt = d(sint)/dt = cost= cos√x

And dt/dx= d√x/dx= 1/2√x

So final answer dy/dx = cos√x/2√x

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ddx[sinx−−√]=cosx−−√⋅ddx[x−−√]=cosx−−√⋅12x−−√

What is the derivative of sin under root x?

Y is equal to root sin x find DY by DX?

What is dy/dx when xy=sin(x+y)?

If sin y = x sin(a+y) , then dy/dx =?

If y = x-1/x, what is dy/dx?

y = sin(√x)

=> y=sin(x^(1/2))

=> y’ = cos(x^(1/2))*d/dx(x^(1/2)

=> y’ =cos(x^½)*½*x^(½-1)

=> y’ = ½*cos(x^½)/x^(½)

=>y’=½{cos(√x)/√x}

dey/dx =cos root x *(1/2 )*x^1/2

dy/dx=C

Answer 2

The value of $\frac{dy}{dx}$  is  $\frac{cos\sqrt{x} }{4\sqrt{x\ sin\sqrt{x} }}$.

Explanation:

Given:

$y=\sqrt{sin\sqrt{x} }$

To Find:

The value of $\frac{dy}{dx}$.

Solution:

As given,$y=\sqrt{sin\sqrt{x} }$.

$\frac{d(\sqrt{sin\sqrt{x} })}{dx}$

$\frac{d f(p)}{dx} =\frac{df}{dp} .\frac{dp}{dx}$

Let $f=\sqrt{p}$  , $p=sin\sqrt{x} }$

        $=\frac{d }{dp}(\sqrt{p} ) .\frac{d}{dx} (sin\sqrt{x} )$

        $=\frac{1}{2\sqrt{p} }.\frac{cos\sqrt{x} }{2\sqrt{x} }$

         $=\frac{1}{2\sqrt{sin\sqrt{x} } }.\frac{cos\sqrt{x} }{2\sqrt{x} }$

          $=\frac{cos\sqrt{x} }{4\sqrt{x\sin\sqrt{x} }}$

Thus,the value of $\frac{dy}{dx}$  is  $\frac{cos\sqrt{x} }{4\sqrt{x\ sin\sqrt{x} }}$.

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