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Question: What is the derivative of \(\sin {{x}^{\cos x}}\) ?...

What is the derivative of sin⁡xcos⁡x\sin {{x}^{\cos x}} ?

Explanation

Solution

We have been asked in the problem to find the differential of a function raised to another function. We will apply the chain rule of differentiation in our problem. The chain rule of differentiation can be given as: \dfrac{d\left[ f\left\\{ g\left( x \right) \right\\} \right]}{dx}=\dfrac{d\left[ f\left\\{ g\left( x \right) \right\\} \right]}{d\left[ g\left( x \right) \right]}\times \dfrac{d\left[ g\left( x \right) \right]}{dx} . We will also make use of the formula for differentiation of a function raised to some power and the differential of some trigonometric identities.

Complete step by step solution:
The expression given to us in our problem is equal to: sin⁡xcos⁡x\sin {{x}^{\cos x}}. Now, let us first assign some terms that we are going to use later in our problem.
Let the given term on which we need to operate a differential be given by ‘y’ . Here, ‘y’ is given to us as:
⇒y=sin⁡xcos⁡x\Rightarrow y=\sin {{x}^{\cos x}}
Then, we need to find the differential of ‘y’ with respect to ‘x’. This can be done as follows:
⇒dydx=d[sin⁡xcos⁡x]dx\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ \sin {{x}^{\cos x}} \right]}{dx}
Applying the chain rule of differentiation on the given expression, our expression can be written as:
⇒dydx=d[sin⁡xcos⁡x]d(cos⁡x)×d(cos⁡x)dx\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ \sin {{x}^{\cos x}} \right]}{d\left( \cos x \right)}\times \dfrac{d\left( \cos x \right)}{dx}
Using the formula for differential of a function raised to some power, that is equal to:
⇒dθndθ=nθn−1\Rightarrow \dfrac{d{{\theta }^{n}}}{d\theta }=n{{\theta }^{n-1}}
We get our equation as:
⇒dydx=cos⁡x.sin⁡x(cos⁡x−1)×d(cos⁡x)dx\Rightarrow \dfrac{dy}{dx}=\cos x.\sin {{x}^{\left( \cos x-1 \right)}}\times \dfrac{d\left( \cos x \right)}{dx}
Now, using the formula for differentiating the cosine function, that is equal to:
⇒d(cos⁡θ)dθ=−sin⁡θ\Rightarrow \dfrac{d\left( \cos \theta \right)}{d\theta }=-\sin \theta
Our equation can be further simplified into:
⇒dydx=cos⁡x.sin⁡x(cos⁡x−1)×(−sin⁡x) ⇒dydx=−cos⁡x.sin⁡x(cos⁡x−1+1) ⇒dydx=−cos⁡x.sin⁡xcos⁡x \begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\cos x.\sin {{x}^{\left( \cos x-1 \right)}}\times \left( -\sin x \right) \\\ & \Rightarrow \dfrac{dy}{dx}=-\cos x.\sin {{x}^{\left( \cos x-1+1 \right)}} \\\ & \Rightarrow \dfrac{dy}{dx}=-\cos x.\sin {{x}^{\cos x}} \\\ \end{aligned}
Thus, our final expression comes out to be −cos⁡x.sin⁡xcos⁡x-\cos x.\sin {{x}^{\cos x}}.
Hence, the derivative of, sin⁡xcos⁡x\sin {{x}^{\cos x}} comes out to be −cos⁡x.sin⁡xcos⁡x-\cos x.\sin {{x}^{\cos x}}.

Note: While applying the chain rule, we should always make sure we are differentiating our fundamental equation with respect to a correct function. Finding out this function is the key to our solution. Also, while applying different formulas simultaneously in one equation, we should make sure that the formulas and calculations are correct.