Question

Find dy/dx if y= root tan root x

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Answer 1

y=√(tan√x)

y=(tan√x)^-1/2

dy/dx=d/dx(tan√x)^-1/2

=1/2√(tan√x).d/dx(tan√x)

=1/2√(tan√x). sec²√x. d√x/dx

=1/2√(tan√x). sec²√x .1/2√x

=(sec²√x)/4√(xtan√x)

Answer 2

The value of dy/dx is $\dfrac{\sec ^2\left(\sqrt{x}\right)}{4\sqrt{x}\sqrt{\tan \left(\sqrt{x}\right)}}$.

Step-by-step explanation:

The given function is

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

We need to find the value of dy/dx.

Differentiate with respect to x.

$\dfrac{dy}{dx}=\frac{1}{2\sqrt{\tan \left(\sqrt{x}\right)}}\cdot \dfrac{d}{dx}\tan \left(\sqrt{x}\right)$                   $[\because \dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}]$

$\dfrac{dy}{dx}=\frac{1}{2\sqrt{\tan \left(\sqrt{x}\right)}}\cdot \sec^2 \left(\sqrt{x}\right)\cdot \dfrac{d}{dx}\sqrt{x}$                $[\because \dfrac{d}{dx}\tan x=\sec^2x]$

$\dfrac{dy}{dx}=\frac{1}{2\sqrt{\tan \left(\sqrt{x}\right)}}\cdot \sec^2 \left(\sqrt{x}\right)\cdot \dfrac{1}{2\sqrt{x}}$                $[\because \dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}]$

$\dfrac{dy}{dx}=\dfrac{\sec ^2\left(\sqrt{x}\right)}{4\sqrt{x}\sqrt{\tan \left(\sqrt{x}\right)}}$

Therefore, the value of dy/dx is $\dfrac{\sec ^2\left(\sqrt{x}\right)}{4\sqrt{x}\sqrt{\tan \left(\sqrt{x}\right)}}$.

#Learn more

If y=sin root x, what is dy/dx?

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