differnetiate using chain rule:- √sin√x
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Answer:
[tex]\large\boxed{\sf{\dfrac{ \cot( \sqrt{x} ) }{4 \sqrt{x} }}}[/tex]
Step-by-step explanation:
[tex]y = \sqrt{ \sin( \sqrt{x} ) } [/tex]
Let's assume,
[tex] t = \sqrt{x } [/tex]
Differentiating both sides,
[tex] = > \frac{dt}{dx} = \frac{1}{2 \sqrt{x} } [/tex]
Also, let's assume,
[tex]u = \sin( \sqrt{x} ) \\ \\ = > u = \sin(t) [/tex]
Differentiating both sides,
[tex] = > \frac{du}{dt} = \cos(t) [/tex]
Therefore, we have,
[tex]y = \sqrt{u} [/tex]
Differentiating both sides,
[tex] = > \frac{dy}{du} = \frac{1}{2 \sqrt{u} } [/tex]
Therefore, we have,
[tex] = > \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dt} \times \frac{dt}{dx} \\ \\ = > \frac{dy}{dx} = \frac{1}{2 \sqrt{u} } \times \cos(t) \times \frac{1}{2 \sqrt{x} } \\ \\ = > \frac{dy}{dx} = \frac{ \cos(t) }{4 \sqrt{ux} } \\ \\ = > \frac{dy}{dx} = \frac{ \cos( \sqrt{x} ) }{4 \sqrt{x} \sin( \sqrt{x} ) } \\ \\ = > \frac{dy}{dx} = \frac{ \cot( \sqrt{x} ) }{4 \sqrt{x} } [/tex]
Answer:
4 square root of x
Step-by-step explanation: