Question

How do you find the derivative of ${{\cos }^{3}}u$ ?

Answer 1

Hint: We recall the definition of composite function$gof\left( u \right)=g\left( f\left( u \right) \right)$. We recall the chain rule of differentiation $\dfrac{dy}{du}=\dfrac{dy}{dv}\times \dfrac{dv}{du}$ where $y=gof={{\cos }^{3}}u$ and $v=f\left( u \right)=\cos u$. We first find $v=f\left( u \right)$ as the function inside the bracket and $y$ as the given function and then differentiate using chain rule and standard derivative of cosine function $\dfrac{d}{dt}\left( \cos t \right)=-\sin t$ \[\]

Complete step-by-step answer:
We know from calculus that the derivative of a function of a real variable measures the rate of change of the functional value with respect to argument or input value. The process of finding derivative is called differentiation. If $f\left( x \right)$ is real valued function then we use the differential operator $\dfrac{d}{dx}$ and find the derivative as
\[\dfrac{d}{dx}f\left( x \right)={{f}^{'}}\left( x \right)\]

If the functions $f\left( u \right),g\left( u \right)$ are real valued defined within sets $f:A\to B$ and $g:B\to C$ then the composite function from A to C is defend as $g\left( f\left( x \right) \right)$ within sets$gof:A\to C$. If we denote $g\left( f\left( u \right) \right)=y$ and $f\left( u \right)=v$ then we can differentiate the composite function using chain rule as
\[\dfrac{d}{dx}g\left( f\left( u \right) \right)=\dfrac{dy}{du}=\dfrac{dy}{dv}\times \dfrac{dv}{du}\]
We are asked to differentiate the function${{\cos }^{3}}u$. We see that it is a composite function which made by functions polynomial cubic function that is ${{u}^{3}}$ and trigonometric function that is ${{\cos }^{3}}u={{\left( \cos u \right)}^{3}}$. Let us assign the function within the bracket as $f\left( u \right)=\cos u=v$ and $g\left( u \right)={{u}^{3}}$. So we have$g\left( f\left( u \right) \right)=g\left( \cos u \right)={{\left( \cos u \right)}^{3}}=y$. We differentiate using chain rule to have;
\[\begin{align}
  & \dfrac{dy}{du}=\dfrac{dy}{dv}\times \dfrac{dv}{du} \\
 & \Rightarrow \dfrac{d}{du}y=\dfrac{d}{dv}y\times \dfrac{d}{du}v \\
 & \Rightarrow \dfrac{d}{du}{{\left( \cos u \right)}^{3}}=\dfrac{d}{d\left( \cos u \right)}{{\left( \cos u \right)}^{3}}\times \dfrac{d}{dx}\left( \cos u \right) \\
\end{align}\]
We know that from standard differentiation of polynomial function as $\dfrac{d}{dt}{{t}^{n}}=n{{t}^{n-1}}$ where $n$ is any real number. We use it for $t=\cos u,n=3$ in the above step to have
\[\Rightarrow \dfrac{d}{du}{{\left( \cos u \right)}^{3}}=3{{\left( \cos u \right)}^{2}}\times \dfrac{d}{dx}\left( \cos u \right)\]
We know that from standard differentiation of cosine function $\dfrac{d}{dt}\cos t=-\sin t$. We use it for $t=u$ in the above step to have
\[\begin{align}
  & \Rightarrow \dfrac{d}{du}{{\left( \cos u \right)}^{3}}=3{{\left( \cos u \right)}^{2}}\times \left( -\sin u \right) \\
 & \Rightarrow \dfrac{d}{du}{{\cos }^{3}}u=-3\sin u {{\cos }^{2}}u \\
\end{align}\]

Note: We must be careful to remember the negative sign in differentiation of $\cos x$ that is $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$ and note that there is no negative sign in differentiation of $\sin x$ that is $\dfrac{d}{dx}\left( \sin x \right)=\cos x$. We can also use the identity ${{\cos }^{3}}x=\dfrac{3\cos x+\cos 3x}{4}$ to differentiate the right hand side but then we have to simplify using $\sin 3x=3\sin x-4{{\sin }^{3}}x$ which will be tedious.