Question

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

Answer 1

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,

& \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{aligned}$$ Hence, the derivative of $$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’.