Question

What is the derivative of this function $f(x) = \sin \left( {\dfrac{1}{{{x^2}}}} \right)?$

Answer 1

Hint: In this question, we have to find the derivative of the given function. We can see that we also have one trigonometric ratio in the equation. So we will use the derivative of the trigonometric function as we will use the formula of the chain rule to solve this question. We will also use the basic power rule .
Formula used:
The chain rule states that:
$\dfrac{d}{{dx}}[f(g(x))] = f'g(x)g'(x)$
The power rule of differentiation is as follows:
$\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$

Complete step by step answer:
In this question, we have
$f(x) = \sin \left( {\dfrac{1}{{{x^2}}}} \right)$
We should note that we can also write
$ \Rightarrow \sin \left( {\dfrac{1}{{{x^2}}}} \right) = \sin ({x^{ - 2}})$
Therefore we have a new expression which is of the form:
$ = \dfrac{d}{{dx}}\sin ({x^{ - 2}})$
Now we know that the derivative of the sine function is cosine function i.e.
$ \Rightarrow \dfrac{d}{{dx}}\sin = \cos $
So by applying the chain rule, we can write the expression as:
$ = \cos ({x^{ - 2}})\left( {\dfrac{d}{{dx}}{x^{ - 2}}} \right)$
Now we will apply the power rule in the second part of the expression, So by comparing from the product rule, we have:
$n = - 2$
Therefore by applying this formula in the equation, we can write:
$ \Rightarrow {x^{ - 2}} = 2{x^{ - 2 - 1}}$
Now by substituting these values in the equation we can write:
$ = \cos ({x^{ - 2}})( - 2{x^{ - 3}})$
Again we can write the negative value from the exponent to the fraction form, so we have:
$ = \dfrac{{ - 2\cos \left( {\dfrac{1}{{{x^2}}}} \right)}}{{{x^3}}}$
Hence the required answer is $\dfrac{{ - 2\cos \left( {\dfrac{1}{{{x^2}}}} \right)}}{{{x^3}}}$ .

Note: Before solving this kind of question, we should always be aware of the basic formula of differentiation. We should note that chain rule is used to find the derivative of the composite functions which means that when we have two functions such as $\left( {{x^2} + 1} \right),\sin 2x...$ and so on. In the above question also we have two functions.