Question

How do you find the derivative of ${\cos ^2}\left( {3x} \right)$ ?

Answer 1

In order to find the derivative of the given expression, we use the chain rule method. We substitute $\cos \left( {3x} \right) = u$, and differentiate both sides with respect to $x$ . Then we take $y = {u^2}$ and differentiate both sides with respect to $u$. We multiply both of them to get our required answer.

Complete step-by-step solution:
The given expression is ${\cos ^2}\left( {3x} \right)$
Now, we need to differentiate the above given expression. In order to do so, we follow the chain rule.
The chain rule is as follows: $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}....\left( {\text{A}} \right)$
Let us call this as equation (A)
Let us take$\cos \left( {3x} \right) = u$. Now let us differentiate both the sides with respect to $x$
$\Rightarrow \dfrac{{du}}{{dx}} = - 3\sin \left( {3x} \right)$
As we know that the differentiation of $\dfrac{d}{{dx}}\cos x = - \sin x$
Now let us take $y = {u^2}$
On differentiating both sides with respect to $u$, we get:
$\Rightarrow \dfrac{{dy}}{{du}} = 2u$
Now, as we know that $u = \cos \left( {3x} \right)$
Therefore placing this value in $2u$
Therefore, $2u = 2\cos \left( {3x} \right)$
Now, if we look back at equation (A), we find that:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$
Thus, $\dfrac{{dy}}{{dx}} = 2u \times \left( { - 3\sin \left( {3x} \right)} \right)$
$\Rightarrow \;\dfrac{{dy}}{{dx}} = \left( {2\cos \left( {3x} \right)} \right) \times \left( { - 3\sin \left( {3x} \right)} \right)$
On further simplifying, we get:
$\Rightarrow \;\dfrac{{dy}}{{dx}} = - 6\cos \left( {3x} \right)\sin \left( {3x} \right)$

Therefore $- 6\cos \left( {3x} \right)\sin \left( {3x} \right)$ is the required answer.

Note: Derivatives are a fundamental tool of Calculus, which is a branch of Mathematics. Derivation or differentiation is a way of finding the instantaneous rate of change of a function based on one of its variables.
We can find the derivative of trigonometric functions, logarithmic functions, etc using different formulas.
The inverse of differentiation is integration.
While differentiation means to break down a function into parts, integration means to add up the infinitesimal parts to give a whole function.
Some common formulas of differentiation are:
$\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$
$\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}$
$\dfrac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\dfrac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$