Question

Differentiate $ \sqrt{\sin \sqrt{x}} $ w.r.t. x.

Answer 1

Hint:Make use of the chain rule of derivatives for composite functions: $

\dfrac{d}{dx}f[g(x)] $ = $

\dfrac{d}{d\ g(x)}f[g(x)] $ × $ \dfrac{d}{dx}g(x) $ .

Also recall the chain rule of derivatives for change of variables: $ \dfrac{dy}{dx} $ = $ \dfrac{dy}{dt} $ × $
\dfrac{dt}{dx} $ .

Substitute $ \sqrt{x} $ = t to avoid confusion. What will be the relation between $ \dfrac{dy}{dx} $ and $
\dfrac{dy}{dt} $ after this substitution?
Recall that $ \dfrac{d}{dx}{{x}^{n}} $ = $ n{{x}^{n−1}} $ and $ \dfrac{d}{dx}\sin x $ = cos x.

Complete step by step solution:

Say the given function is y = $ \sqrt{\sin \sqrt{x}} $ .
Let's substitute $ \sqrt{x} $ = t.

On differentiating it with respect to x, we get:
$ \dfrac{d}{dx}{{x}^{\dfrac{1}{2}}} $ = $ \dfrac{d}{dx}t $
⇒ $ \dfrac{1}{2}{{x}^{\dfrac{1}{2}−1}} $ = $ \dfrac{dt}{dx} $
⇒ $ \dfrac{1}{2\sqrt{x}} $ = $ \dfrac{dt}{dx} $ ... (1)

Now, after the substitution, the given function is y = $ \sqrt{\sin t} $ . Differentiating it with respect to t, and using the chain rule, we get:

$ \dfrac{dy}{dt} $ = $ \dfrac{dy}{d(\sin t)}\sqrt{\sin t} $ × $ \dfrac{d}{dt}(\sin t) $
⇒ $ \dfrac{dy}{dt} $ = $ \dfrac{1}{2\sqrt{\sin t}} $ × cos t

Back substituting $ \sqrt{x} $ = t, we get:
⇒ $ \dfrac{dy}{dt} $ = $ \dfrac{1}{2\sqrt{\sin \sqrt{x}}} $ × $ \cos \sqrt{x} $ ... (2)

Now, using the chain rule of derivatives for change of variables, we have:
$ \dfrac{dy}{dx} $ = $ \dfrac{dy}{dt} $ × $ \dfrac{dt}{dx} $

Using the values in equations (1) and (2), we get:
⇒ $ \dfrac{dy}{dt} $ = $ \dfrac{1}{2\sqrt{\sin \sqrt{x}}} $ × $ \cos \sqrt{x} $ × $ \dfrac{2\sqrt{x}}{1} $

⇒ $ \dfrac{dy}{dt} $ = $ \dfrac{\sqrt{x}\cos \sqrt{x}}{\sqrt{\sin \sqrt{x}}} $ , which is the required derivative with respect to x.

Note:We can try differentiating without substituting also, but it will be a little bit cumbersome writing it. The product rule of derivatives: $ \dfrac{d}{dx}[f(x)g(x)] $ = $ f(x)\left[ \dfrac{d}{dx}g(x) \right] $ + $ \left[
\dfrac{d}{dx}f(x) \right]g(x) $ .

The derivatives of sin x and cos x are found using the first principle (definition) of derivatives. The first principle states that the derivative of a function f(x) is: $ \dfrac{d}{dx}f(x) $ = $ \underset{h\to
0}{\mathop{\lim }}\,\dfrac{f(x+h)−f(x)}{h} $ . The derivative of a function is the rate of change of the value of the function with respect to the change in the value of the independent variable x. The derivative of a function y = f(x) is often also denoted by y' = f'(x).