
How do you find the derivative of $y = \cos 2x$ ?
Answer
548.4k+ views
Hint: Here we have a function inside another function, so we have to make use of chain rule to find the derivative of the given function. The chain rule is given by: $\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)$. Substitute $f(x)$ as $\cos 2x$ and $g(x)$ as $2x$ and simplify to get the required answer.
Complete step-by-step answer:
Whenever they give a function first see whether a single function is given or function inside a function, because if we have a single function then we can directly differentiate the function, if function inside another function is given then we have to make use of chain rule to solve the problem.
Here we have a function $2x$ inside the other function $\cos $ .
So, now we make use of chain rule to solve the problem. The chain rule is given by:
$\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)$
Which is nothing but first we have to differentiate the outer function and then the inside function.
We know that the derivative of $\cos x$ is nothing but $ - \sin x$ , so by using this in chain rule we can differentiate the given function that is $y = \cos 2x$. Now, if we differentiate $\cos 2x$ we get $ - \sin 2x$ but again we need to differentiate the term $2x$ to get the final answer, which can be done in following way,
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\cos 2x} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \sin 2x.\dfrac{d}{{dx}}\left( {2x} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \sin 2x.2$ or we can also write it as $ \Rightarrow \dfrac{{dy}}{{dx}} = - 2\sin 2x$
Therefore the derivative of the given function $y = \cos 2x$ is $\dfrac{{dy}}{{dx}} = - 2\sin 2x$ .
Note: Whenever we have to find derivatives of given functions you should be knowing the basic trigonometric derivatives so that we can solve in the correct way, otherwise we may end up in the wrong answer. In case of function inside function do not forget to make use of chain rule.
Complete step-by-step answer:
Whenever they give a function first see whether a single function is given or function inside a function, because if we have a single function then we can directly differentiate the function, if function inside another function is given then we have to make use of chain rule to solve the problem.
Here we have a function $2x$ inside the other function $\cos $ .
So, now we make use of chain rule to solve the problem. The chain rule is given by:
$\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)$
Which is nothing but first we have to differentiate the outer function and then the inside function.
We know that the derivative of $\cos x$ is nothing but $ - \sin x$ , so by using this in chain rule we can differentiate the given function that is $y = \cos 2x$. Now, if we differentiate $\cos 2x$ we get $ - \sin 2x$ but again we need to differentiate the term $2x$ to get the final answer, which can be done in following way,
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\cos 2x} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \sin 2x.\dfrac{d}{{dx}}\left( {2x} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \sin 2x.2$ or we can also write it as $ \Rightarrow \dfrac{{dy}}{{dx}} = - 2\sin 2x$
Therefore the derivative of the given function $y = \cos 2x$ is $\dfrac{{dy}}{{dx}} = - 2\sin 2x$ .
Note: Whenever we have to find derivatives of given functions you should be knowing the basic trigonometric derivatives so that we can solve in the correct way, otherwise we may end up in the wrong answer. In case of function inside function do not forget to make use of chain rule.
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