Question

How do you differentiate \[f\left( x \right)=\sin \left( \dfrac{\pi }{x} \right)\]?

Answer 1

Hint: Use the chain rule of differentiation to find the derivative of the given function \[\sin \left( \dfrac{\pi }{x} \right)\]. First differentiate \[\sin \left( \dfrac{\pi }{x} \right)\] with respect to \[\left( \dfrac{\pi }{x} \right)\] and then differentiate \[\left( \dfrac{\pi }{x} \right)\] with respect to x and take their product to get the answer. Use the basic differentiation formula: - \[\dfrac{d\sin x}{dx}=\cos x\] and \[d\left( \dfrac{1}{x} \right)=\dfrac{-1}{{{x}^{2}}}\] to get the required derivative.

Complete step by step answer:
Here, we have been provided with a function \[\sin \left( \dfrac{\pi }{x} \right)\] and we have to find its derivative. That means we have to differentiate this function.
Now, let us assume the given function as f (x). So, we have,
\[\Rightarrow f\left( x \right)=\sin \left( \dfrac{\pi }{x} \right)\]
Since, the variable is x therefore we have to find the derivative with respect to x, i.e., \[\dfrac{df\left( x \right)}{dx}\]. So, differentiating both the sides, we get,
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=\dfrac{d\sin \left( \dfrac{\pi }{x} \right)}{dx}\]
Using the chain rule of differentiation, we get,
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=\dfrac{d\sin \left( \dfrac{\pi }{x} \right)}{dx}\times \dfrac{d\left( \dfrac{\pi }{x} \right)}{dx}\]
What we are doing is, we are first differentiating f (x) with respect to \[\left( \dfrac{\pi }{x} \right)\] and then we are differentiating \[\left( \dfrac{\pi }{x} \right)\] with respect to x and then finally taking their product. Now, we know that the derivative of sine function is a cosine function, i.e., \[\dfrac{d\sin x}{dx}=\cos x\], so we have,
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=\cos \left( \dfrac{\pi }{x} \right)\times \dfrac{d\left( \dfrac{\pi }{x} \right)}{dx}\]
Since, \[\pi \] is a constant therefore it can be taken out of the derivative. So, we get,
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=\cos \left( \dfrac{\pi }{x} \right)\times \pi \left[ \dfrac{d\left( \dfrac{1}{x} \right)}{dx} \right]\]
We can write \[\dfrac{1}{x}\] as \[{{x}^{-1}}\], so we have,
\[\Rightarrow \dfrac{df\left( x \right)}{dx}=\cos \left( \dfrac{\pi }{x} \right)\times \pi \left[ \dfrac{d\left( {{x}^{-1}} \right)}{dx} \right]\]
Using the formula: - \[\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}\], we get,
\[\begin{align}
  & \Rightarrow \dfrac{df\left( x \right)}{dx}=\cos \left( \dfrac{\pi }{x} \right)\times \pi \left[ \left( -1 \right)\times {{x}^{-1-1}} \right] \\
 & \Rightarrow \dfrac{df\left( x \right)}{dx}=-\pi \cos \left( \dfrac{\pi }{x} \right){{x}^{-2}} \\
 & \Rightarrow \dfrac{df\left( x \right)}{dx}=\dfrac{-\pi \cos \left( \dfrac{\pi }{x} \right)}{{{x}^{2}}} \\
\end{align}\]
Hence, the derivative of \[f\left( x \right)=\sin \left( \dfrac{\pi }{x} \right)\] is \[\dfrac{-\pi \cos \left( \dfrac{\pi }{x} \right)}{{{x}^{2}}}\].

Note:
One may note that we are assuming the angle \[\left( \dfrac{\pi }{x} \right)\] in radian. Sometimes we will be provided with the angel in degrees like \[\cos {{\left( 180x \right)}^{\circ }}\]. In such cases we cannot differentiate the function directly. First, we have to convert the angle in radians using the formula: - \[{{1}^{\circ }}=\dfrac{\pi }{180}\] radian and then only we can differentiate. You must remember all the basic formulas of differentiation like: - the chain rule, product rule, \[\dfrac{u}{v}\] rule etc. because they are frequently used in the topic ‘calculus’.